A lead ball of mass 2 kg moving with a veloc | vedclass

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(B) According to the law of conservation of linear momentum:$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$(2 \ kg)(1.5 \ ms^{-1}) + (3 \ kg)(0) = (2 \ kg)(v

Vedclass · Accepted Answer

$0.75$

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(B) According to the law of conservation of linear momentum:$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$(2 \ kg)(1.5 \ ms^{-1}) + (3 \ kg)(0) = (2 \ kg)(v_1) + (3 \ kg)(1 \ ms^{-1})$$3 = 2v_1 + 3$$2v_1 = 0 \implies v_1 = 0 \ ms^{-1}$Initial Kinetic Energy $(KE_i)$:$KE_i = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} 	imes 2 	imes (1.5)^2 + 0 = 2.25 \ J$Final Kinetic Energy $(KE_f)$:$KE_f = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} 	imes 2 	imes (0)^2 + \frac{1}{2} 	imes 3 	imes (1)^2 = 1.5 \ J$Change in Kinetic Energy $(\Delta KE)$:$\Delta KE = KE_i - KE_f = 2.25 \ J - 1.5 \ J = 0.75 \ J$

$A$ lead ball of mass $2 \ kg$ moving with a velocity of $1.5 \ ms^{-1}$ collides with a ball of mass $3 \ kg$ at rest. After the collision,the second ball moves with a velocity of $1 \ ms^{-1}$ in the original direction of motion of the first ball. Find $\Delta KE$ in $J$.

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