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(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

Respuesta :

Answer:

Speed, v₂ = 8 m/s

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 :

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]

[tex]0=4\times (-14)+7\times v_2[/tex]

[tex]v_2=-8\ m/s[/tex]

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.

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