A body of mass 10 , kg moving with a speed of | vedclass

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(B) According to the law of conservation of linear momentum,the total momentum before the collision is equal to the total momentum after the collision

Vedclass · Accepted Answer

$30$

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(B) According to the law of conservation of linear momentum,the total momentum before the collision is equal to the total momentum after the collision.Let $m_1 = 10 \, kg$,$u_1 = 3 \, m/s$,$m_2 = 5 \, kg$,and $u_2 = 0 \, m/s$.Let $V$ be the final velocity of the combined mass $(m_1 + m_2) = 15 \, kg$.$m_1 u_1 + m_2 u_2 = (m_1 + m_2) V$$10 	imes 3 + 5 	imes 0 = (10 + 5) V$$30 = 15 V$$V = 2 \, m/s$Now,the kinetic energy $(KE)$ of the composite mass is given by:$KE = \frac{1}{2} (m_1 + m_2) V^2$$KE = \frac{1}{2} 	imes 15 	imes (2)^2$$KE = \frac{1}{2} 	imes 15 	imes 4$$KE = 30 \, J$.

$A$ body of mass $10 \, kg$ moving with a speed of $3 \, m/s$ collides with another stationary body of mass $5 \, kg$. As a result,the two bodies stick together. The $KE$ of the composite mass will be .......... $J$.

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