Ball A of mass 1 ,kg moving along a straight | vedclass

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Question

(A) For ball $A$: $m_A = 1 \,kg$, $u_A = 4 \,ms^{-1}$. For ball $B$: $m_B = 3 \,kg$, $u_B = 0 \,ms^{-1}$.Total initial momentum $p_i = m_A u_A + m_B u

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(A) For ball $A$: $m_A = 1 \,kg$, $u_A = 4 \,ms^{-1}$. For ball $B$: $m_B = 3 \,kg$, $u_B = 0 \,ms^{-1}$.Total initial momentum $p_i = m_A u_A + m_B u_B = (1 	imes 4) + (3 	imes 0) = 4 \,kg \cdot ms^{-1}$.After collision, both bodies stick together and move with a common velocity $v$.Total final momentum $p_f = (m_A + m_B)v = (1 + 3)v = 4v$.By the law of conservation of linear momentum, $p_i = p_f$, so $4 = 4v$, which gives $v = 1 \,ms^{-1}$.The force exerted on ball $B$ is the rate of change of momentum of ball $B$:$F_B = \frac{\Delta p_B}{\Delta t} = \frac{m_B(v - u_B)}{\Delta t} = \frac{3(1 - 0)}{0.1} = \frac{3}{0.1} = 30 \,N$.

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