Question

audit of mass 10 gram moving with a velocity of hundred metre per second strikes a wooden block of mass 990 gram and gets embedded in it find the velocity with which combined mass would move also find the loss in kinetic energy during the collision ​

Answer 1

Answer:

The two blocks will move together with speed v = 1 m/s and the total loss of kinetic energy in this collision is 49.5 J

Explanation:

When audit of mass 10 gm strike the wooden block of mass 990 gm and get stick to it

So we can use momentum conservation for this

$m_1v_1 = (m_1 + m_2) v$

so we have

$10(100) = (10 + 990) v$

$v = 1 m/s$

So the final speed of the system is 1 m/s

Now loss in kinetic energy is given as

$\Delta E = \frac{1}{2}m_1 v_1^2 - \frac{1}{2}(m_1 + m_2)v^2$

$\Delta E = \frac{1}{2}(0.01)(100)^2 - \frac{1}{2}1(1^2)$

$\Delta E = 49.5 J$

#Learn

Topic : Collision

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