Two skaters P and Q are skating towards each | vedclass

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Question

(A) Let the initial distance between $P$ and $Q$ be $x$.Case $(I)$: $P$ runs towards $Q$ at $v_P = 1 \,ms^{-1}$,$Q$ is stationary $(v_Q = 0)$. The bal

Vedclass · Accepted Answer

One every $2.5 \,s$ in case $(I)$ and one every $3.3 \,s$ in case $(II)$

Vedclass · Answer

(A) Let the initial distance between $P$ and $Q$ be $x$.Case $(I)$: $P$ runs towards $Q$ at $v_P = 1 \,ms^{-1}$,$Q$ is stationary $(v_Q = 0)$. The ball is thrown with speed $v_b = 2 \,ms^{-1}$ relative to the ground.The time taken for the first ball to reach $Q$ is $t_1 = \frac{x}{v_b} = \frac{x}{2}$.The second ball is thrown at $t = 5 \,s$. At this time,$P$ has moved a distance $d = v_P 	imes 5 = 1 	imes 5 = 5 \,m$ towards $Q$. The new distance between $P$ and $Q$ is $(x - 5)$.The time taken for the second ball to reach $Q$ from the moment it is thrown is $t' = \frac{x - 5}{2}$.So,the second ball reaches $Q$ at time $t_2 = 5 + t' = 5 + \frac{x - 5}{2} = 5 + \frac{x}{2} - 2.5 = \frac{x}{2} + 2.5$.The interval between receiving the balls is $\Delta t = t_2 - t_1 = (\frac{x}{2} + 2.5) - \frac{x}{2} = 2.5 \,s$.Case $(II)$: $Q$ runs towards $P$ at $v_Q = 1 \,ms^{-1}$,$P$ is stationary $(v_P = 0)$. The ball is thrown with speed $v_b = 2 \,ms^{-1}$ relative to the ground.The first ball reaches $Q$ at time $t_1$ when the distance covered by the ball and $Q$ equals $x$: $v_b t_1 + v_Q t_1 = x \implies (2 + 1)t_1 = x \implies t_1 = \frac{x}{3}$.The second ball is thrown at $t = 5 \,s$. At this time,$Q$ has moved a distance $d = v_Q 	imes 5 = 1 	imes 5 = 5 \,m$ towards $P$. The new distance between $P$ and $Q$ is $(x - 5)$.The time taken for the second ball to reach $Q$ from the moment it is thrown is $t'$ such that $(v_b + v_Q)t' = x - 5 \implies (2 + 1)t' = x - 5 \implies t' = \frac{x - 5}{3}$.So,the second ball reaches $Q$ at time $t_2 = 5 + t' = 5 + \frac{x - 5}{3} = 5 + \frac{x}{3} - \frac{5}{3} = \frac{x}{3} + \frac{10}{3}$.The interval between receiving the balls is $\Delta t = t_2 - t_1 = (\frac{x}{3} + \frac{10}{3}) - \frac{x}{3} = \frac{10}{3} \,s \approx 3.33 \,s$.

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