Chapter 9 Force And Laws Of Motion Solutions
Question - 21 : - A hockey ball of mass 200 g travelling at 10ms–1 is struck by a hockey stick so as to return it along itsoriginal path with a velocity at 5 ms–1. Calculate the magnitude ofchange of momentum occurred in the motion of the hockey ball by the forceapplied by the hockey stick.
Answer - 21 : -
Given, mass of theball (m) = 200g
Initial velocity ofthe ball (u) = 10 m/s
Final velocity of theball (v) = – 5m/s
Initial momentum ofthe ball = mu = 200g × 10 ms-1 = 2000 g.m.s-1
Final momentum of theball = mv = 200g × –5 ms-1 = –1000 g.m.s-1
Therefore, the changein momentum (mv – mu) = –1000 g.m.s-1 – 2000 g.m.s-1 =–3000 g.m.s-1
This implies that themomentum of the ball reduces by 1000 g.m.s-1 after being struckby the hockey stick.
Question - 22 : - A bullet of mass 10 g travelling horizontally with a velocity of 150 m s–1 strikesa stationary wooden block and comes to rest in 0.03 s. Calculate the distanceof penetration of the bullet into the block. Also calculate the magnitude ofthe force exerted by the wooden block on the bullet.
Answer - 22 : -
Given, mass of thebullet (m) = 10g (or 0.01 kg)
Initial velocity ofthe bullet (u) = 150 m/s
Terminal velocity ofthe bullet (v) = 0 m/s
Time period (t) = 0.03s
To find the distanceof penetration, the acceleration of the bullet must be calculated.
Therefore, forceexerted by the wooden block on the bullet (F) = 0.01kg × (-5000 ms-2)
= -50 N
This implies that thewooden block exerts a force of magnitude 50 N on the bullet in the directionthat is opposite to the trajectory of the bullet.
Question - 23 : - An object of mass 1 kg travelling in astraight line with a velocity of 10 ms–1 collides with, andsticks to, a stationary wooden block of mass 5 kg. Then they both move offtogether in the same straight line. Calculate the total momentum just beforethe impact and just after the impact. Also, calculate the velocity of thecombined object.
Answer - 23 : -
Given, mass of theobject (m1) = 1kg
Mass of the block (m2)= 5kg
Initial velocity ofthe object (u1) = 10 m/s
Initial velocity ofthe block (u2) = 0
Mass of the resultingobject = m1 + m2 = 6kg
Velocity of theresulting object (v) =?
Total momentum beforethe collision = m1u1 + m2u2 =(1kg) × (10m/s) + 0 = 10 kg.m.s-1
As per the law ofconservation of momentum, the total momentum before the collision is equal tothe total momentum post the collision. Therefore, the total momentum post thecollision is also 10 kg.m.s-1
Now, (m1 +m2) × v = 10kg.m.s-1
The resulting objectmoves with a velocity of 1.66 meters per second.
Question - 24 : - An object of mass 100 kg is accelerateduniformly from a velocity of 5 ms–1 to 8 ms–1 in6 s. Calculate the initial and final momentum of the object. Also, find themagnitude of the force exerted on the object.
Answer - 24 : -
Given, mass of theobject (m) = 100kg
Initial velocity (u) =5 m/s
Terminal velocity (v)= 8 m/s
Time period (t) = 6s
Now, initial momentum(m × u) = 100kg × 5m/s = 500 kg.m.s-1
Final momentum (m × v)= 100kg × 8m/s = 800 kg.m.s-1
Therefore, the objectaccelerates at 0.5 ms-2. This implies that the force acting on theobject (F = ma) is equal to:
F = (100kg) × (0.5 ms-2)= 50 N
Therefore, a force of 50 Nis applied on the 100kg object, which accelerates it by 0.5 ms-2.
Question - 25 : - Akhtar, Kiran, and Rahul were riding in a motorcar that was moving with ahigh velocity on an expressway when an insect hit the windshield and got stuckon the windscreen. Akhtar and Kiran started pondering over the situation. Kiransuggested that the insect suffered a greater change in momentum as compared tothe change in momentum of the motorcar (because the change in the velocity ofthe insect was much more than that of the motorcar). Akhtar said that since themotorcar was moving with a larger velocity, it exerted a larger force on theinsect. And as a result the insect died. Rahul while putting an entirely newexplanation said that both the motorcar and the insect experienced the sameforce and a change in their momentum. Comment on these suggestions.
Answer - 25 : -
As per the law ofconservation of momentum, the total momentum before the collision between theinsect and the car is equal to the total momentum after the collision.Therefore, the change in the momentum of the insect is much greater than thechange in momentum of the car (since force is proportional to mass).
Akhtar’s assumption ispartially right. Since the mass of the car is very high, the force exerted onthe insect during the collision is also very high.
Kiran’s statement isfalse. The change in momentum of the insect and the motorcar is equal byconservation of momentum. The velocity of insect changes accordingly due to itsmass as it is very small compared to the motorcar. Similarly, the velocity ofmotorcar is very insignificant because its mass is very large compared to theinsect.
Rahul’s statement iscompletely right. As per the third law of motion, the force exerted by theinsect on the car is equal and opposite to the force exerted by the car on theinsect. However, Rahul’s suggestion that the change in the momentum is the samecontradicts the law of conservation of momentum.
Question - 26 : - How much momentum will a dumb-bell of mass 10 kg transfer to the floor ifit falls from a height of 80 cm? Take its downward acceleration to be 10 ms–2.
Answer - 26 : -
Given, mass of thedumb-bell (m) = 10kg
Distance covered (s) =80cm = 0.8m
Initial velocity (u) =0 (it is dropped from a position of rest)
Acceleration (a) =10ms-2
Terminal velocity (v)=?
Momentum of thedumb-bell when it hits the ground = mv
v = 4 m/s
The momentumtransferred by the dumb-bell to the floor = (10kg) × (4 m/s) = 40 kg.m.s-1
Question - 27 : - The following is the distance-time table of anobject in motion: | Time (seconds) | Distance (meters) |
| 0 | 0 |
| 1 | 1 |
| 2 | 8 |
| 3 | 27 |
| 4 | 64 |
| 5 | 125 |
| 6 | 216 |
| 7 | 343 |
(a)What conclusion can you draw about the acceleration? Is it constant,increasing, decreasing, or zero? (b) What do you infer about the forces actingon the object?Answer - 27 : -
(a) The distancecovered by the object at any time interval is greater than any of the distancescovered in previous time intervals. Therefore, the acceleration of the objectis increasing.
(b) As per the secondlaw of motion, force = mass × acceleration. Since the mass of the objectremains constant, the increasing acceleration implies that the force acting onthe object is increasing as well
Question - 28 : - Two persons manage to push a motorcar of mass 1200 kg at a uniformvelocity along a level road. The same motorcar can be pushed by three personsto produce an acceleration of 0.2 m s-2. With what force does eachperson push the motorcar? (Assume that all persons push the motorcar with thesame muscular effort)
Answer - 28 : -
Given, mass of the car(m) = 1200kg
When the third personstarts pushing the car, the acceleration (a) is 0.2ms-2. Therefore,the force applied by the third person (F = ma) is given by:
F = 1200kg × 0.2 ms-2 =240N
The force applied bythe third person on the car is 240 N. Since all 3 people push with the samemuscular effort, the force applied by each person on the car is 240 N.
Question - 29 : - A hammer of mass 500 g, moving at 50 m s-1, strikes a nail. The nailstops the hammer in a very short time of 0.01 s. What is the force of the nailon the hammer?
Answer - 29 : -
Given, mass of thehammer (m) = 500g = 0.5kg
Initial velocity of thehammer (u) = 50 m/s
Terminal velocity ofthe hammer (v) = 0 (the hammer is stopped and reaches a position of rest).
Time period (t) =0.01s
a = -5000ms-2
Therefore, the forceexerted by the hammer on the nail (F = ma) can be calculated as:
F = (0.5kg) * (-5000ms-2) = -2500 N
As per the third lawof motion, the nail exerts an equal and opposite force on the hammer. Since theforce exerted on the nail by the hammer is -2500 N, the force exerted on thehammer by the nail will be +2500 N.
Question - 30 : - A motorcar of mass 1200 kg is moving along astraight line with a uniform velocity of 90 km/h. Its velocity is slowed downto 18 km/h in 4 s by an unbalanced external force. Calculate the accelerationand change in momentum. Also calculate the magnitude of the force required.
Answer - 30 : -
Given, mass of the car(m) = 1200kg
Initial velocity (u) =90 km/hour = 25 meters/sec
Terminal velocity (v)= 18 km/hour = 5 meters/sec
Time period (t) = 4seconds
Therefore, theacceleration of the car is -5 ms-2.
Initial momentum ofthe car = m × u = (1200kg) × (25m/s) = 30,000 kg.m.s-1
Final momentum of thecar = m × v = (1200kg) × (5m/s) = 6,000 kg.m.s-1
Therefore, change inmomentum (final momentum – initial momentum) = (6,000 – 30,000) kg.m.s-1
= -24,000 kg.m.s-1
External force applied= mass of car × acceleration = (1200kg) × (-5 ms-2) = -6000N
Therefore, themagnitude of force required to slow down the vehicle to 18 km/hour is 6000 N