Two identical balls P and Q moving in the x-y | vedclass

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(C) Since the balls are identical,let their mass be $m$. According to the law of conservation of linear momentum,the total momentum before the collisi

Vedclass · Accepted Answer

$v_x = 1 \ m/s, v_y = 2 \ m/s$

Vedclass · Answer

(C) Since the balls are identical,let their mass be $m$. According to the law of conservation of linear momentum,the total momentum before the collision must equal the total momentum after the collision. Initial momentum in $x$-direction: $P_{ix} = m(v_{Px}) + m(v_{Qx}) = m(6) + m(-5) = m \ kg \ m/s$. Initial momentum in $y$-direction: $P_{iy} = m(v_{Py}) + m(v_{Qy}) = m(0) + m(2) = 2m \ kg \ m/s$. After the collision,ball $P$ comes to rest,so $v'_{Px} = 0$ and $v'_{Py} = 0$. Let the velocity components of ball $Q$ after the collision be $v'_{Qx}$ and $v'_{Qy}$. Final momentum in $x$-direction: $P_{fx} = m(0) + m(v'_{Qx}) = m(v'_{Qx})$. Final momentum in $y$-direction: $P_{fy} = m(0) + m(v'_{Qy}) = m(v'_{Qy})$. Equating initial and final momenta: $m(v'_{Qx}) = m \implies v'_{Qx} = 1 \ m/s$. $m(v'_{Qy}) = 2m \implies v'_{Qy} = 2 \ m/s$. Therefore,the velocity components of ball $Q$ after the collision are $v_x = 1 \ m/s$ and $v_y = 2 \ m/s$.

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