## Laws of Motion Questions and Answers Part-4

1. A parachutist of weight ‘w’ strikes the ground with his legs fixed and comes to rest with an upward acceleration of magnitude 3 g. Force exerted on him by ground during landing is
a) w
b) 2w
c) 3w
d) 4w

Explanation: Resultant force is w + 3w = 4w

2. At a place where the acceleration due to gravity is $10m sec^{-2}$  a force of 5 kg-wt acts on a body of mass 10 kg initially at rest. The velocity of the body after 4 second is
a) 5 $m sec^{-1}$
b) 10 $m sec^{-1}$
c) 20 $m sec^{-1}$
d) 50 $m sec^{-1}$

Explanation: 3. In a rocket of mass 1000 kg fuel is consumed at a rate of 40 kg/s. The velocity of the gases ejected from the rocket is $5\times10^{4}m\diagup s$  . The thrust on the rocket is
a) $2\times10^{3}N$
b) $5\times10^{4}N$
c) $2\times10^{6}N$
d) $2\times10^{9}N$

Explanation: 4. A man is standing on a weighing machine placed in a lift. When stationary his weight is recorded as 40 kg. If the lift is accelerated upwards with an acceleration of $2m\diagup sec^{2}$ , then the weight recorded in the machine will be $\left(g=10m\diagup sec^{2}\right)$
a) 32 kg
b) 40 kg
c) 42 kg
d) 48 kg

Explanation: In stationary lift man weighs 40 kg i.e. 400 N.
When lift accelerates upward it's apparent weight = m(g + a) = 40(10 + 2) = 480 N i.e. 48 kg
For the clarity of concepts in this problem kg-wt can be used in place of kg

5. A body of mass 4 kg weighs 4.8 kg when suspended in a moving lift. The acceleration of the lift is
a) 9.8 $m sec^{-2}$  downwards
b) 9.8 $m sec^{-2}$  upwards
c) 1.96 $m sec^{-2}$  downwards
d) 1.96 $m sec^{-2}$  upwards

Explanation: As the apparent weight increase therefore we can say that acceleration of the lift is in upward direction.
R = m(g + a)
4.8 g = 4(g + a)
a = 0.2g = 1.96 m /s2

6. An elevator weighing 6000 kg is pulled upward by a cable with an acceleration of $5m s^{-2}$ . Taking g to be $10m s^{-2}$ , then the tension in the cable is
a) 6000 N
b) 9000 N
c) 60000 N
d) 90000 N

Explanation: T = m(g + a) = 6000 (10 + 5) = 90000 N

7. A vehicle of 100 kg is moving with a velocity of 5 m/sec. To stop it in $\frac{1}{10}sec$ , the required force in opposite direction is
a) 5000 N
b) 500 N
c) 50 N
d) 1000 N

Explanation: 8. A boy having a mass equal to 40 kilograms is standing in an elevator. The force felt by the feet of the boy will be greatest when the elevator $\left(g=9.8 metres\diagup sec^{2}\right)$
a) Stands still
b) Moves downward at a constant velocity of 4 metres/sec
c) Accelerates downward with an acceleration equal to 4 $metres\diagup sec^{2}$
d) Accelerates upward with an acceleration equal to 4 $metres\diagup sec^{2}$

Explanation: Accelerates upward with an acceleration equal to 4 $metres\diagup sec^{2}$

9. A rocket has an initial mass of $20\times 10^{3}kg$  . If it is to blast off with an initial acceleration of $4 ms^{-2}$ , the initial thrust needed is $\left(g\cong10 ms^{-2}\right)$
a) $6\times 10^{4}N$
b) $28\times 10^{4}N$
c) $20\times 10^{4}N$
d) $12\times 10^{4}N$ a) $\frac{3}{2}g$
b) $\frac{g}{3}$
c) $\frac{2}{3}g$ 