Particle A moving with a velocity v = 10 m/s | vedclass

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(A) Let the mass of both particles be $m$. Let $V_1$ and $V_2$ be the velocities of particles $A$ and $B$ after the collision,respectively.By the law

Vedclass · Accepted Answer

$9.95$

Vedclass · Answer

(A) Let the mass of both particles be $m$. Let $V_1$ and $V_2$ be the velocities of particles $A$ and $B$ after the collision,respectively.By the law of conservation of momentum:$m 	imes 10 + 0 = m V_1 + m V_2 \Rightarrow V_1 + V_2 = 10$ ... $(i)$Given that the kinetic energy of the system decreases by $1 \%$,the final kinetic energy $K_f = 0.99 K_i$.$\frac{1}{2} m V_1^2 + \frac{1}{2} m V_2^2 = 0.99 	imes (\frac{1}{2} m 	imes 10^2)$$V_1^2 + V_2^2 = 0.99 	imes 100 = 99$ ... (ii)We know that $(V_1 + V_2)^2 = V_1^2 + V_2^2 + 2 V_1 V_2$.Substituting the values: $10^2 = 99 + 2 V_1 V_2 \Rightarrow 100 = 99 + 2 V_1 V_2 \Rightarrow 2 V_1 V_2 = 1 \Rightarrow V_1 V_2 = 0.5$.Now,$(V_1 - V_2)^2 = (V_1 + V_2)^2 - 4 V_1 V_2 = 100 - 4(0.5) = 100 - 2 = 98$.$V_1 - V_2 = \sqrt{98} = 7\sqrt{2} \approx 9.899 \ m/s$.Adding $(i)$ and (ii): $2 V_1 = 10 + 9.899 = 19.899 \Rightarrow V_1 \approx 9.95 \ m/s$.

Particle $A$ moving with a velocity $v = 10 \ m/s$ experiences a head-on collision with a stationary particle $B$ of the same mass. As a result of the collision,the kinetic energy of the system decreases by $1 \%$. The speed of particle $A$ after the collision is: (in $m/s$)

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