When the carpet is beaten, it is suddenly set into motion. The dust particles tend to remain at rest due to inertia of rest, therefore the dust comes out of it.

When a bust starts suddenly, the lower part of the luggage kept on the roof being in contact with the bus begins to move forward with the speed of bus, but the upper part tends to remain at rest due to inertia of rest. Therefore, the upper part is left behind and hence luggage falls backward. So, it is advised to tie any luggage kept on the roof of a bus with a rope.

A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because
(a) the batsman did not hit the ball hard enough.
(b) velocity is proportional to the force exerted on the ball.
(c) there is a force on the ball opposing the motion.
(d) there is no unbalanced force on the ball, so the ball would want to come to rest.

Answer

The ball slows down and comes to rest due to opposing forces of air resistance and frictional force on the ball opposing its motion. Therefore the choice (c) there is a force on the ball opposing the motion is correct.

Initial velocity, u = 0 
Distance travelled, s = 400 m
Time taken, t = 20 s

We know, s = ut + ½ at²

Or, 400 = 0 + ½ a (20)²
Or, a = 2 ms^–2
Now, m = 7 MT = 7000 kg, a = 2 ms^–2
Or, F = ma = 7000 × 2 = 14000 N Ans.

Initial velocity of the stone, u= 20 m/s
Final velocity of the stone, v= 0
Distance covered by the stone, s= 50 m
Since, v² - u²
2 = 2as,
Or, 0 - 202 = 2a × 50,
Or, a = – 4 ms^-2
Force of friction, F = ma = – 4N

A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate:
(a) the net accelerating force;
(b) the acceleration of the train; and
(c)the force of wagon 1 on wagon 2.

Answer

(a) Force exerted by the engine, F = 40000 N
Frictional force offered by the track, F_f = 5000 N 
Net accelerating force, Fa = F − F_f = 40000 − 5000 = 35000 N
Hence, the net accelerating force is 35000 N.

(b) Acceleration of the train = a
The engine exerts a force of 40000 N on all the five wagons. 
Net accelerating force on the wagons, Fa = 35000 N
Mass of the wagons, m = Mass of a wagon x Number of wagons
Mass of a wagon = 2000 kg
Number of wagons = 5
∴ m = 2000 × 5 = 10000 kg
Total mass, M = m  = 10000 kg
From Newton’s second law of motion:
Fa= Ma
a=Fam   = 35000 10000    = 3.5 ms^-2
Hence, the acceleration of the wagons and the train is 3.5 m/s².

(c) Mass of all the wagons except wagon 1 is 4 × 2000 = 8000 kg
Acceleration of the wagons = 3.5 m/s²
Thus, force exerted on all the wagons except wagon 1 
= 8000 × 3.5 = 28000 N
Therefore, the force exerted by wagon 1 on the remaining four wagons is 28000 ​ N.
Hence, the force exerted by wagon 1 on wagon 2 is 28000 ​ N.

Mass of the automobile vehicle, m= 1500 kg 
Final velocity, v= 0 (finally the automobile stops)
Acceleration of the automobile, a= −1.7 ms⁻²
From Newton’s second law of motion:
Force = Mass x Acceleration = 1500 x (−1.7) = −2550 N
Hence, the force between the automobile and the road is −2550 N, in the direction opposite to the motion of the automobile.

What is the momentum of an object of mass m, moving with a velocity v?
(a)    (mv)2    (b)    mv2    (c)    ½ mv2    (d)    mv

Answer

(d) mv
Mass of the object = m
Velocity = v
Momentum = Mass x Velocity 
Momentum = mv