In the previous section we completed the discussion on Newton's Third law of motion. We also saw the Law of conservation of Momentum. In this section we will see some more solved examples.
Solved example 2.13
From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m s-1 . Calculate the initial recoil velocity of the rifle.
Solution:
1. Masses:
• Given: Mass of bullet = mB = 50 g = 0.05 kg
• Given: Mass of rifle = mR = 4 kg
2. Initial velocities:
• The bullet accelerates from rest. So uB = 0
• The rifle accelerates from rest. So uR = 0
3. Final velocities:
• Given that velocity of bullet = vB= 35 m s-1.
• We have to find vR.
4. Total momentum before the pistol is fired = (mBuB +mRuR) = (0.05 × 0 + 4 × 0) = 0 kg m s-1
• Total momentum after the pistol is fired = (mBvB +mRvR) = (0.05 × 35 + 4 × vR) = (4vR+ 1.75) kg m s-1
5. According to the law of conservation of momentum, Total momentum after the fire = Total momentum before the fire. So we can write:
4vR+ 1.75 = 0 ⇒ vR = - 1.75⁄4 = -0.44 m s-1.
(The negative sign indicates that, the direction of motion of the rifle is opposite to the direction of motion of the bullet)
Solved example 2.14
Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 m s-1 and 1 m s-1 respectively. They collide and after the collision, the first object moves at a velocity of 1.67 m s-1. Determine the velocity of the second object.
Solution:
1. Masses:
• Given: Mass of first object = mA = 100 g = 0.1 kg
• Given: Mass of second object = mB = 200g = 0.2 kg
2. Initial velocities:
• Given: Initial velocity of first object = uA = 2 m s-1
• Given: Initial velocity of second object = uB = 1 m s-1
Note that both velocities are taken as positive because, they travel in the same direction.
3. Final velocities:
• Given: Final velocity of first object = vA = 1.67 m s-1
• We have to find vB.
4. Total momentum before the collision = (mAuA +mBuB) = (0.1 × 2 + 0.2 × 1) = 0.4 kg m s-1
• Total momentum after the collision = (mAvA +mBvB) = (0.1 × 1.67 + 0.2 × vB) = (0.2vB+ 0.167) kg m s-1
5. According to the law of conservation of momentum, Total momentum after the collision = Total momentum before the collision. So we can write:
0.4 = 0.2vB+ 0.167 ⇒ 0.2vB = 0.233 ⇒ vB = 0.233⁄0.2 = 1.165 m s-1.
(The positive sign of vB indicates that, the direction of motion of the second object is same as the initial direction of motion of the objects)
Solved example 2.15
Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 m s-1 before the collision during which they stick together. What will be the velocity of the combined object after collision?
Solution:
In this problem, we have a special situation:
• The objects stick together after the collision. That means, after the collision, the objects move together as a single mass.
So we will write the steps as follows:
1. Initial Masses:
• Given: Mass of object A = mA = 1.5 kg
• Given: Mass of object B = mB = 1.5 kg
2. Final mass = mf = mA + mB = 1.5 + 1.5 = 3 kg
3. Initial velocities:
• Object A was moving with a velocity. Given uA = 2.5 m s-1
• Object B was moving with a velocity. Given uB = 2.5 m s-1
♦ But object B was moving in a direction opposite to that of object A. So the velocity of object B has to be taken as negative. Thus we can write: uB = -2.5 m s-1
4. Final velocity:
• After collision, the two object move together. We have to find the final velocity vf with which they move together.
5. Total momentum before collision = (mAuA + mBuB) = (1.5 × 2.5 + 1.5 × -2.5) = 3.75 - 3.75 = 0 kg m s-1.
• Total momentum after collision = mfvf = 3 × vf = 3vf kg m s-1
6. According to the law of conservation of momentum, Total momentum after collision = Total momentum before collision. So we can write:
-3vf = 0 ⇒ vf = 0
Solved example 2.16
A hockey ball of mass 200 g travelling at 10 m s-1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 m s-1. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Solution:
1. Mass of the hockey ball = 200 g = 0.2 kg
2. Initial velocity of the hockey ball = 10 m s-1
3. So initial momentum before collision = 0.2 × 10 = 2 kg m s-1
4. Final velocity of the hockey ball = -5 m s-1
Note that the final velocity is given a negative sign because it returns along the same path. That is., it travels in the opposite direction after collision
5. Final momentum after collision = 0.2 × (-5) = -1 kg m s-1 .
6. So difference between the two momenta = 2 - (-1) = 2 + 1 = 3 kg m s-1.
Solved example 2.17
A bullet of mass 10 g travelling horizontally with a velocity of 150 m s-1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.
Solution:
1. Initial velocity with which the bullet hits the wooden block = u = 150 m s-1
2. Final velocity of the bullet in the block = v = 0 m s-1
3. Time in which the bullet comes to rest = t = 0.03 s
4. So the velocity decreases from 150 to zero m s-1. It is clear that there is deceleration (negative acceleration). To find this acceleration 'a', we can use the first equation of motion.
5. We have: v = u + at ⇒ 0 = 150 + a × 0.03 ⇒ a = -150⁄0.03 = -5000 m s-2.
6. We now have the initial velocity u, final velocity v, acceleration a and time of travel t. We want the distance travelled. We can use the third equation of motion. Because it connects all these quantities
7. So we can write:
v2 = u2 + 2as ⇒ 02 = 1502 + 2 × -5000 × s ⇒ 0 = 22500 + -10000s ⇒ s = 2.25 m.
8. The negative acceleration was due to the resistance of the wooden block. That means, the wooden block applied a resisting force F on the bullet. We want the value of this F
9. We know that force = mass × acceleration ⇒ F = ma = 0.01× -5000 = -50 N (∵ mass = 10 gram = 0.01 kg)
10. So we can write:
• The distance of penetration = 2.25 m
• The force exerted by the wooden block on the bullet = 50 N
Solved example 2.18
An object of mass 1 kg travelling in a straight line with a velocity of 10 m s-1 collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.
Solution:
In this problem, we have a special situation:
• The objects stick together after the collision. That means, after the collision, the objects move together as a single mass.
So we will write the steps as follows:
1. Initial Masses:
• Given: Mass of object A = mA = 1.0 kg
• Given: Mass of wooden block = mB = 5.0 kg
2. Final mass = mf = mA + mB = 1 + 5 = 6 kg
3. Initial velocities:
• Object A was moving with a velocity. Given uA = 10 m s-1
• Wooden block was stationary. So uB = 0 m s-1
4. Final velocity:
• After collision, the two object move together. We have to find the final velocity vf with which they move together.
5. Total momentum before collision = (mAuA + mBuB) = (1 × 10 + 5 × 0) = 10 + 0 = 10 kg m s-1.
• Total momentum after collision = mfvf = 6 × vf = 6vf kg m s-1
6. According to the law of conservation of momentum, Total momentum after collision = Total momentum before collision. So we can write:
6vf = 10 ⇒ vf = 1.66 m s-1.
7. So the total momentum before the impact = 10 kg m s-1
• The total momentum after the impact = 6vf = 6 × 1.66 = 9.96 kg m s-1.
• Velocity of the combined object = vf = 1.66 m s-1.
Solved example 2.19
An object of mass 100 kg is accelerated uniformly from a velocity of 5 m s-1 to 8 m s-1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
Solution:
1. Mass of the object = 100 kg
2. Initial velocity of the object = 5 m s-1
3. So initial momentum = 100 × 5 = 500 kg m s-1
4. Final velocity of the object = 8 m s-1
5. So final momentum = 100 × 8 = 800 kg m s-1
6. Time in which the change in velocity took place = t = 6 s
7. So the velocity increases from 5 to 8 m s-1. It is clear that there is acceleration. To find this acceleration 'a', we can use the first equation of motion.
8. We have: v = u + at ⇒ 8 = 5 + a × 6 ⇒ a = 3⁄6 = 0.5 m s-2.
9. We know that force = mass × acceleration ⇒ F = ma = 100 × 0.5 = 50 N
Solved example 2.20
How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 m s-1 .
Solution:
1. Mass of the dumb-bell = m = 10 kg
2. Height of fall = distance of travel = s = 80 cm = 0.8 m
3. Acceleration = a = 10 m s-2.
4. The dumb-bell falls from rest. So it has an initial velocity u = 0
5. We have to find the final velocity v. We can use the third equation of motion. Because it connects all these quantities
6. So we can write:
v2 = u2 + 2as ⇒ v2 = 02 + 2 × 10 × 0.8 ⇒ v2 = 16 ⇒ v = 4 m s-1.
7. Let us analyse the situation:
(i) The dumb-bell falls from a height of 80 cm
(ii) It's velocity goes on increasing because it is acted upon by an acceleration of 10 m s-2.
(iii) When the distance of travel becomes 80 cm, the velocity becomes 4 m s-1.
(iv) The velocity would have increased even more, if it was allowed to travel more distance.
(v) But the travel comes to an abrupt stop. Because it met the floor at 80 cm.
8. The momentum of the dumb-bell at the time of hitting the floor = mv = 10 × 4 = 40 kg m s-1.
9. Now we consider the law of conservation of momentum:
(i) The system consists of two objects: The dumb-bell and the floor
(ii) Momentum of the dumb-bell just before collision = 40 kg m s-1.
(iii) Momentum of the floor just before collision = 0 kg m s-1.
(iv) Total momentum just before collision = 40 + 0 = 40 kg m s-1.
10. This must be the total momentum after collision also. So a momentum of 40 kg m s-1 is transferred to the floor.
Solved example 2.21
Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 m s-2. With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.)
Solution:
1. When two persons push the car, it moves with a uniform velocity. Uniform velocity means no acceleration. That means, no resultant external force is acting on the car. That means, the frictional resistance is just balanced by the effort of two persons.
2. Let the force provided by each person be FP. Then the frictional force FF will be equal to 2FP. So we can write: FF = 2FP
3. Now a third person also puts in the same effort FP. The combined pushing force is greater than the frictional force. That is why, there is a resultant force acting on the car. Because of this resultant force, the car begins to move with acceleration 'a' which is given as 2 m s-2.
4. The resultant force acting on the car will be equal to 3FP - FF.
This resultant force is equal to mass × acceleration = 1200 × 0.2 = 240 N
So we can write: 3FP - FF = 240 N
5. But from (2) we get FF = 2FP. Substituting this in (4) we get:
3FP - 2FP = 240 N ⇒ FP = 240 N
So the force with which each person push the car is 240 N
Solved example 2.22
A hammer of mass 500 g, moving at 50 m s-1, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?
Solution:
1. The hammer is moving with an initial velocity u of 50 m s-1.
2. It is stopped by the nail. The time taken by the nail to stop the hammer is 0.01 s
3. The hammer is 'stopped' by the nail. That means, the final velocity v of the hammer is 0 m s-1
4. Since there is a velocity change, there is acceleration. Here the velocity decreases from 50 m s-1 to zero. So it is deceleration or negative acceleration. We have to find this acceleration first. We can use the first equation of motion:
5. v = u + at ⇒ 0 = 50 + a × 0.01 ⇒ a = -50⁄0.01 = -5000 m s-2.
6. The 500 g mass of the hammer is subjected to an acceleration of -5000 m s-2.
7. So force = mass × acceleration = 0.5 × -5000 = -2500 N
8. This much force is exerted by the hammer on the nail. By newtons third law, the nail exerts the same force in the opposite direction.
9. So the force exerted by the nail on the hammer = + 2500 N
Solved example 2.23
A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 kmph. Its velocity is slowed down to 18 kmph in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.
Solution:
1. Initial velocity u = 90 kmph = 90 × 1000⁄3600 = 25 m s-1.
2. Final velocity v = 18 kmph = 18 × 1000⁄3600 = 5 m s-1.
3. Time duration in which the velocity changed = 4 s
4. To find acceleration a, we can use the first equation of motion:
5. v = u + at ⇒ 5 = 25 + a × 4 ⇒ a = -20⁄4 = -5 m s-2.
6. Change in momentum = m(v-u) = 1200 × (5-25) = 1200 × -20 = -24000 kg m s-1.
7. Force = mass × acceleration = 1200 × -5 = -6000 N. The negative sign indicates that, the force is acting in the direction opposite to the direction of motion.
In the next chapter, we will see Gravitational force.
Solved example 2.13
From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m s-1 . Calculate the initial recoil velocity of the rifle.
Solution:
1. Masses:
• Given: Mass of bullet = mB = 50 g = 0.05 kg
• Given: Mass of rifle = mR = 4 kg
2. Initial velocities:
• The bullet accelerates from rest. So uB = 0
• The rifle accelerates from rest. So uR = 0
3. Final velocities:
• Given that velocity of bullet = vB= 35 m s-1.
• We have to find vR.
4. Total momentum before the pistol is fired = (mBuB +mRuR) = (0.05 × 0 + 4 × 0) = 0 kg m s-1
• Total momentum after the pistol is fired = (mBvB +mRvR) = (0.05 × 35 + 4 × vR) = (4vR+ 1.75) kg m s-1
5. According to the law of conservation of momentum, Total momentum after the fire = Total momentum before the fire. So we can write:
4vR+ 1.75 = 0 ⇒ vR = - 1.75⁄4 = -0.44 m s-1.
(The negative sign indicates that, the direction of motion of the rifle is opposite to the direction of motion of the bullet)
Solved example 2.14
Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 m s-1 and 1 m s-1 respectively. They collide and after the collision, the first object moves at a velocity of 1.67 m s-1. Determine the velocity of the second object.
Solution:
1. Masses:
• Given: Mass of first object = mA = 100 g = 0.1 kg
• Given: Mass of second object = mB = 200g = 0.2 kg
2. Initial velocities:
• Given: Initial velocity of first object = uA = 2 m s-1
• Given: Initial velocity of second object = uB = 1 m s-1
Note that both velocities are taken as positive because, they travel in the same direction.
3. Final velocities:
• Given: Final velocity of first object = vA = 1.67 m s-1
• We have to find vB.
4. Total momentum before the collision = (mAuA +mBuB) = (0.1 × 2 + 0.2 × 1) = 0.4 kg m s-1
• Total momentum after the collision = (mAvA +mBvB) = (0.1 × 1.67 + 0.2 × vB) = (0.2vB+ 0.167) kg m s-1
5. According to the law of conservation of momentum, Total momentum after the collision = Total momentum before the collision. So we can write:
0.4 = 0.2vB+ 0.167 ⇒ 0.2vB = 0.233 ⇒ vB = 0.233⁄0.2 = 1.165 m s-1.
(The positive sign of vB indicates that, the direction of motion of the second object is same as the initial direction of motion of the objects)
Solved example 2.15
Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 m s-1 before the collision during which they stick together. What will be the velocity of the combined object after collision?
Solution:
In this problem, we have a special situation:
• The objects stick together after the collision. That means, after the collision, the objects move together as a single mass.
So we will write the steps as follows:
1. Initial Masses:
• Given: Mass of object A = mA = 1.5 kg
• Given: Mass of object B = mB = 1.5 kg
2. Final mass = mf = mA + mB = 1.5 + 1.5 = 3 kg
3. Initial velocities:
• Object A was moving with a velocity. Given uA = 2.5 m s-1
• Object B was moving with a velocity. Given uB = 2.5 m s-1
♦ But object B was moving in a direction opposite to that of object A. So the velocity of object B has to be taken as negative. Thus we can write: uB = -2.5 m s-1
4. Final velocity:
• After collision, the two object move together. We have to find the final velocity vf with which they move together.
5. Total momentum before collision = (mAuA + mBuB) = (1.5 × 2.5 + 1.5 × -2.5) = 3.75 - 3.75 = 0 kg m s-1.
• Total momentum after collision = mfvf = 3 × vf = 3vf kg m s-1
6. According to the law of conservation of momentum, Total momentum after collision = Total momentum before collision. So we can write:
-3vf = 0 ⇒ vf = 0
Solved example 2.16
A hockey ball of mass 200 g travelling at 10 m s-1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 m s-1. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Solution:
1. Mass of the hockey ball = 200 g = 0.2 kg
2. Initial velocity of the hockey ball = 10 m s-1
3. So initial momentum before collision = 0.2 × 10 = 2 kg m s-1
4. Final velocity of the hockey ball = -5 m s-1
Note that the final velocity is given a negative sign because it returns along the same path. That is., it travels in the opposite direction after collision
5. Final momentum after collision = 0.2 × (-5) = -1 kg m s-1 .
6. So difference between the two momenta = 2 - (-1) = 2 + 1 = 3 kg m s-1.
Solved example 2.17
A bullet of mass 10 g travelling horizontally with a velocity of 150 m s-1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.
Solution:
1. Initial velocity with which the bullet hits the wooden block = u = 150 m s-1
2. Final velocity of the bullet in the block = v = 0 m s-1
3. Time in which the bullet comes to rest = t = 0.03 s
4. So the velocity decreases from 150 to zero m s-1. It is clear that there is deceleration (negative acceleration). To find this acceleration 'a', we can use the first equation of motion.
5. We have: v = u + at ⇒ 0 = 150 + a × 0.03 ⇒ a = -150⁄0.03 = -5000 m s-2.
6. We now have the initial velocity u, final velocity v, acceleration a and time of travel t. We want the distance travelled. We can use the third equation of motion. Because it connects all these quantities
7. So we can write:
v2 = u2 + 2as ⇒ 02 = 1502 + 2 × -5000 × s ⇒ 0 = 22500 + -10000s ⇒ s = 2.25 m.
8. The negative acceleration was due to the resistance of the wooden block. That means, the wooden block applied a resisting force F on the bullet. We want the value of this F
9. We know that force = mass × acceleration ⇒ F = ma = 0.01× -5000 = -50 N (∵ mass = 10 gram = 0.01 kg)
10. So we can write:
• The distance of penetration = 2.25 m
• The force exerted by the wooden block on the bullet = 50 N
Solved example 2.18
An object of mass 1 kg travelling in a straight line with a velocity of 10 m s-1 collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.
Solution:
In this problem, we have a special situation:
• The objects stick together after the collision. That means, after the collision, the objects move together as a single mass.
So we will write the steps as follows:
1. Initial Masses:
• Given: Mass of object A = mA = 1.0 kg
• Given: Mass of wooden block = mB = 5.0 kg
2. Final mass = mf = mA + mB = 1 + 5 = 6 kg
3. Initial velocities:
• Object A was moving with a velocity. Given uA = 10 m s-1
• Wooden block was stationary. So uB = 0 m s-1
4. Final velocity:
• After collision, the two object move together. We have to find the final velocity vf with which they move together.
5. Total momentum before collision = (mAuA + mBuB) = (1 × 10 + 5 × 0) = 10 + 0 = 10 kg m s-1.
• Total momentum after collision = mfvf = 6 × vf = 6vf kg m s-1
6. According to the law of conservation of momentum, Total momentum after collision = Total momentum before collision. So we can write:
6vf = 10 ⇒ vf = 1.66 m s-1.
7. So the total momentum before the impact = 10 kg m s-1
• The total momentum after the impact = 6vf = 6 × 1.66 = 9.96 kg m s-1.
• Velocity of the combined object = vf = 1.66 m s-1.
Solved example 2.19
An object of mass 100 kg is accelerated uniformly from a velocity of 5 m s-1 to 8 m s-1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
Solution:
1. Mass of the object = 100 kg
2. Initial velocity of the object = 5 m s-1
3. So initial momentum = 100 × 5 = 500 kg m s-1
4. Final velocity of the object = 8 m s-1
5. So final momentum = 100 × 8 = 800 kg m s-1
6. Time in which the change in velocity took place = t = 6 s
7. So the velocity increases from 5 to 8 m s-1. It is clear that there is acceleration. To find this acceleration 'a', we can use the first equation of motion.
8. We have: v = u + at ⇒ 8 = 5 + a × 6 ⇒ a = 3⁄6 = 0.5 m s-2.
9. We know that force = mass × acceleration ⇒ F = ma = 100 × 0.5 = 50 N
Solved example 2.20
How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 m s-1 .
Solution:
1. Mass of the dumb-bell = m = 10 kg
2. Height of fall = distance of travel = s = 80 cm = 0.8 m
3. Acceleration = a = 10 m s-2.
4. The dumb-bell falls from rest. So it has an initial velocity u = 0
5. We have to find the final velocity v. We can use the third equation of motion. Because it connects all these quantities
6. So we can write:
v2 = u2 + 2as ⇒ v2 = 02 + 2 × 10 × 0.8 ⇒ v2 = 16 ⇒ v = 4 m s-1.
7. Let us analyse the situation:
(i) The dumb-bell falls from a height of 80 cm
(ii) It's velocity goes on increasing because it is acted upon by an acceleration of 10 m s-2.
(iii) When the distance of travel becomes 80 cm, the velocity becomes 4 m s-1.
(iv) The velocity would have increased even more, if it was allowed to travel more distance.
(v) But the travel comes to an abrupt stop. Because it met the floor at 80 cm.
8. The momentum of the dumb-bell at the time of hitting the floor = mv = 10 × 4 = 40 kg m s-1.
9. Now we consider the law of conservation of momentum:
(i) The system consists of two objects: The dumb-bell and the floor
(ii) Momentum of the dumb-bell just before collision = 40 kg m s-1.
(iii) Momentum of the floor just before collision = 0 kg m s-1.
(iv) Total momentum just before collision = 40 + 0 = 40 kg m s-1.
10. This must be the total momentum after collision also. So a momentum of 40 kg m s-1 is transferred to the floor.
Solved example 2.21
Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 m s-2. With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.)
Solution:
1. When two persons push the car, it moves with a uniform velocity. Uniform velocity means no acceleration. That means, no resultant external force is acting on the car. That means, the frictional resistance is just balanced by the effort of two persons.
2. Let the force provided by each person be FP. Then the frictional force FF will be equal to 2FP. So we can write: FF = 2FP
3. Now a third person also puts in the same effort FP. The combined pushing force is greater than the frictional force. That is why, there is a resultant force acting on the car. Because of this resultant force, the car begins to move with acceleration 'a' which is given as 2 m s-2.
4. The resultant force acting on the car will be equal to 3FP - FF.
This resultant force is equal to mass × acceleration = 1200 × 0.2 = 240 N
So we can write: 3FP - FF = 240 N
5. But from (2) we get FF = 2FP. Substituting this in (4) we get:
3FP - 2FP = 240 N ⇒ FP = 240 N
So the force with which each person push the car is 240 N
Solved example 2.22
A hammer of mass 500 g, moving at 50 m s-1, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?
Solution:
1. The hammer is moving with an initial velocity u of 50 m s-1.
2. It is stopped by the nail. The time taken by the nail to stop the hammer is 0.01 s
3. The hammer is 'stopped' by the nail. That means, the final velocity v of the hammer is 0 m s-1
4. Since there is a velocity change, there is acceleration. Here the velocity decreases from 50 m s-1 to zero. So it is deceleration or negative acceleration. We have to find this acceleration first. We can use the first equation of motion:
5. v = u + at ⇒ 0 = 50 + a × 0.01 ⇒ a = -50⁄0.01 = -5000 m s-2.
6. The 500 g mass of the hammer is subjected to an acceleration of -5000 m s-2.
7. So force = mass × acceleration = 0.5 × -5000 = -2500 N
8. This much force is exerted by the hammer on the nail. By newtons third law, the nail exerts the same force in the opposite direction.
9. So the force exerted by the nail on the hammer = + 2500 N
Solved example 2.23
A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 kmph. Its velocity is slowed down to 18 kmph in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.
Solution:
1. Initial velocity u = 90 kmph = 90 × 1000⁄3600 = 25 m s-1.
2. Final velocity v = 18 kmph = 18 × 1000⁄3600 = 5 m s-1.
3. Time duration in which the velocity changed = 4 s
4. To find acceleration a, we can use the first equation of motion:
5. v = u + at ⇒ 5 = 25 + a × 4 ⇒ a = -20⁄4 = -5 m s-2.
6. Change in momentum = m(v-u) = 1200 × (5-25) = 1200 × -20 = -24000 kg m s-1.
7. Force = mass × acceleration = 1200 × -5 = -6000 N. The negative sign indicates that, the force is acting in the direction opposite to the direction of motion.
In the next chapter, we will see Gravitational force.
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