Question

The engine of an unpowered toy train is rolling at a constant speed on a level track, as
shown in Figure. The engine collides with a stationary toy truck, and joins with it. (3)
Before the collision, the toy engine is travelling at 0.35m/ s. The mass of the engine is
0.40kg. The mass of the truck is 0.11kg.
(a) What is the condition for the momentum of an isolated system to remain conserved?
(b) Calculate the momentum of the toy engine before the collision.
(c) Calculate the speed of the joined engine and truck immediately after the collision.

Answer 1

Answer:

a) mometum before collision must be equal to the momentum after collision in a closed system.

b) 0.40* 0.35 = 0.14 kgm/s

c) 0.35*0.40 + 0.35*0.11 = ( 0.40+0.11)*v

   0.1785/(0.40+0.11) = v

    therefore, v= 0.35 m/s

Explanation:

here, v is the final velocity of the engine and the truck,

momentum = mass*velocity is used.

here momentum before the collision must be equal to the momentum after the collision.

as they are joined together after the collision, they move at the same final velocity.

hence, the momentum of the engine before collision+ momentum of the truck before collision= momentum of truck and engine moving together after the collision.

m1u1+ m2u2=(m1+m2)*v

pardon me if the answer is wrong or you might have done this in another way. Thank you.