Question

A car of mass 1,200 kg is moving with a uniform speed of 10 m/s. At point P, it increases its speed to 15 m/s and moves with a constant speed. At point Q, the car throws out some objects of a total mass of 200 kg and maintains the previous speed.
​

Answer 1

Given info : A car of mass 1,200 kg is moving with a uniform speed of 10 m/s. At point P, it increases its speed to 15 m/s and moves with a constant speed. At point Q, the car throws out some objects of a total mass of 200 kg and maintains the previous speed.

To find :

1. kinetic energy of the car is ..
2. the momentum of the car is ..

solution :

1. before point P,

kinetic energy of the car = $\frac{1}{2}mv^2=\frac{1}{2}\times1200kg\times(10m/s)^2$

= 60000 J

hence, option (B) is incorrect choice.

after point Q,

kinetic energy of the car = $\frac{1}{2}$ × (1200 kg - 200 kg ) × ( 15 m/s)²

= 112500 J

hence, option (A) is incorrect choice.

between points P and Q ,

kinetic energy of the car = $\frac{1}{2}$ × 1200 kg × (15 m/s)²

= 135000 J

hence option (d) is incorrect choice.

between points P and Q speed of the car remains constant and that is 15 m/s. so the kinetic energy between them remains constant.

hence, option (C) is correct choice.

2. before point P,

momentum of the car = mu = 1200 kg × 10 m/s = 12000 kg m/s

hence, option (A) is incorrect choice.

after point Q,

momentum of the car = m'v = (1200 kg - 200 kg) × 15 m/s = 15000 kgm/s

hence, option (B) is incorrect choice.

momentum of the car  before point P ≠ momentum of the car after point Q

hence, option (c) is incorrect choice.

at any point between points P and Q = mv = 1200 kg × 15 m/s  = 18000 kgm/s

hence, option (d) is correct choice.