Question

(33-34) A horizontal spring is compressed and a 4 kg block is placed on the left of the spring and a 7 kg block on the right. After the system is released from rest the 4 kg block slides to the left at 14 m/s and the 7 kg block slides to the right. The surface is frictionless. 33. The speed of the 7 kg block after it loses contact with the spring is a) 4 m/s b) 6 m/s c) 8 m/s d) 10 m/s e) 12 m/s

Answer 1

Speed, v₂ = 8 m/s

Step-by-step explanation:

It is given that,

Mass of first block, m₁ = 4 kg

Mass of second block m₂ = 7 kg

Initially, the system is at rest, u₁ = u₂ = 0

Velocity of 4 kg block, v₁ = -14 m/s

Let v₂ is the velocity of the 7 kg block after it loses contact with the spring. Using the conservation of momentum as :

`m_1u_1+m_2u_2=m_1v_1+m_2v_2`

`0=4* (-14)+7* v_2`

`v_2=-8\ m/s`

So, the speed of the 7 kg block after it loses contact with the spring is 8 m/s. Hence, this is the required solution.