A particle of mass 1, kg moving with velocity | vedclass

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(B) Let the initial velocity of the $1\, kg$ particle be $u_1 = 1\, m/s$ and its final velocity be $v_1$. The initial kinetic energy is $K_i = \frac{1

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$\frac{1}{3}\, kg$

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(B) Let the initial velocity of the $1\, kg$ particle be $u_1 = 1\, m/s$ and its final velocity be $v_1$. The initial kinetic energy is $K_i = \frac{1}{2} 	imes 1 	imes (1)^2 = 0.5\, J$.Since the particle loses $\frac{3}{4}$ of its kinetic energy,the final kinetic energy is $K_f = K_i - \frac{3}{4}K_i = \frac{1}{4}K_i = \frac{1}{4} 	imes 0.5 = 0.125\, J$.Thus,$\frac{1}{2} 	imes 1 	imes v_1^2 = 0.125 \Rightarrow v_1^2 = 0.25 \Rightarrow v_1 = \pm 0.5\, m/s$.Assuming the particle continues in the same direction,$v_1 = 0.5\, m/s$.Using conservation of linear momentum: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$.$1(1) + m(0) = 1(0.5) + m v_2 \Rightarrow m v_2 = 0.5$.Using the coefficient of restitution for an elastic collision $(e=1)$: $v_2 - v_1 = u_1 - u_2$.$v_2 - 0.5 = 1 - 0 \Rightarrow v_2 = 1.5\, m/s$.Substituting $v_2$ into the momentum equation: $m(1.5) = 0.5 \Rightarrow m = \frac{0.5}{1.5} = \frac{1}{3}\, kg$.

$A$ particle of mass $1\, kg$ moving with velocity $1\, m/s$,collides elastically with another particle of mass $m$ initially at rest. In the collision,the particle of mass $1\, kg$ loses $\frac{3}{4}$ of its kinetic energy $(K.E.)$. The value of $m$ is:

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