A metal ball of mass 2 ,kg moving with a spee | vedclass

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Question

$(A)$ According to the law of conservation of momentum, the total momentum before the collision equals the total momentum after the collision.
$m_1 u

Vedclass · Accepted Answer

$60$

Vedclass · Answer

$(A)$ According to the law of conservation of momentum, the total momentum before the collision equals the total momentum after the collision.
$m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$
Given $m_1 = 2 \,kg$, $u_1 = 10 \,ms^{-1}$, $m_2 = 3 \,kg$, and $u_2 = 0 \,ms^{-1}$.
$2 	imes 10 + 3 	imes 0 = (2 + 3)v$
$20 = 5v \implies v = 4 \,ms^{-1}$.
Initial kinetic energy $(K_i)$ = $\frac{1}{2} m_1 u_1^2 = \frac{1}{2} 	imes 2 	imes (10)^2 = 100 \,J$.
Final kinetic energy $(K_f)$ = $\frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} 	imes (2 + 3) 	imes (4)^2 = \frac{1}{2} 	imes 5 	imes 16 = 40 \,J$.
Loss in kinetic energy = $K_i - K_f = 100 \,J - 40 \,J = 60 \,J$.

$A$ metal ball of mass $2 \,kg$ moving with a speed of $10 \,ms^{-1}$ has a head-on collision with a stationary ball of mass $3 \,kg$. If after the collision, both the balls move together, then the loss in kinetic energy due to the collision is: (in $\,J.$)

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