Question

A 4.0 kg mass is released from rest at the top of a frictionless ramp 2.0 m above ground. It slides down the ramp to a level, horizontal portion of frictionless track and then collides with a stationary 8.0 kg mass. Immediately after the collision, the larger mass has a velocity v_l = +3.5 m/s (i.e., heading to the right).

The velocity v_s of the smaller mass just after the collision:

A) -0.74 m/s B) +0.74 m/s C) -2.6 m/s D) +2.6 m/s E) 0 m/s

Answer 1

(A)−0.74m/s

Step-by-step explanation:

We are given that

Mass ,
`m_1`=4 kg

Distance, d=2 m

`m_2=8 kg`

`v_2=+3.5 m/s`

`u_2=0`

We have to find the velocity of smaller mass just after the collision.

Initial velocity of smaller mass,u=0

`u_1=√(2gd)=√(2* 9.8* 2)=6.26m/s`

Velocity of smaller mass at the bottom

According to law of conservation of momentum

`m_1u_1+m_2u_2=m_1v_1+m_2v_2`

`4* 6.26+0=4v_s+8* 3.5`

`4v_s=4* 6.26-8* 3.5`

`v_s=(4* 6.26-8* 3.5)/(4)`

`v_s=-0.74m/s`