A body of mass 1,kg collides head-on elastica | vedclass

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(B) Let $m_1 = 1\,kg$,$m_2 = 3\,kg$,$u_1$ be the initial velocity of the $1\,kg$ mass,and $u_2 = 0$ be the initial velocity of the $3\,kg$ mass.After

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(B) Let $m_1 = 1\,kg$,$m_2 = 3\,kg$,$u_1$ be the initial velocity of the $1\,kg$ mass,and $u_2 = 0$ be the initial velocity of the $3\,kg$ mass.After the collision,the velocity of the $1\,kg$ mass is $v_1 = -2\,m/s$ (as it reverses direction).By the law of conservation of linear momentum:$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$1(u_1) + 3(0) = 1(-2) + 3(v_2)$$u_1 = -2 + 3v_2 \quad \dots(1)$For an elastic collision,the coefficient of restitution $e = 1$:$e = \frac{v_2 - v_1}{u_1 - u_2} = 1$$1 = \frac{v_2 - (-2)}{u_1 - 0} \Rightarrow u_1 = v_2 + 2 \quad \dots(2)$Substituting $v_2 = u_1 - 2$ into equation $(1)$:$u_1 = -2 + 3(u_1 - 2)$$u_1 = -2 + 3u_1 - 6$$2u_1 = 8$$u_1 = 4\,m/s$.

$A$ body of mass $1\,kg$ collides head-on elastically with a stationary body of mass $3\,kg$. After the collision,the smaller body reverses its direction of motion and moves with a speed of $2\,m/s$. The initial speed of the smaller body before the collision is $..........\,m/s$.

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