A mass of 20, kg moving with a speed of 10, m | vedclass

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Question

(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

$800$

Vedclass · Answer

(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 $m_1 = 20\, kg$,$v_1 = 10\, m/s$,$m_2 = 5\, kg$,and $v_2 = 0\, m/s$.$m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_{sys}$$20 	imes 10 + 5 	imes 0 = (20 + 5) v_{sys}$$200 = 25 v_{sys}$$v_{sys} = \frac{200}{25} = 8\, m/s$The kinetic energy of the composite mass is given by $K.E. = \frac{1}{2} (m_1 + m_2) v_{sys}^2$.$K.E. = \frac{1}{2} 	imes (20 + 5) 	imes (8)^2$$K.E. = \frac{1}{2} 	imes 25 	imes 64$$K.E. = 25 	imes 32 = 800\, J$.

$A$ mass of $20\, kg$ moving with a speed of $10\, m/s$ collides with another stationary mass of $5\, kg$. As a result of the collision,the two masses stick together. The kinetic energy of the composite mass will be ............. $J$.

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