Initially, ball 1 rests on an incline of height h , and ball 2 rests on an incli
ID: 1427269 • Letter: I
Question
Initially, ball 1 rests on an incline of height h, and ball 2 rests on an incline of height h/2 as shown in the figure below. They are released from rest simultaneously and collide in the trough of the track. If
m2 = 9m1
and the collision is elastic, find an expression for the velocity of each ball immediately after the collision. (Assume the balls slide but do not roll. Use the following as necessary: g and h. Due to the nature of this problem, do not use rounded intermediate values in your calculations—including answers submitted in WebAssign.)
Explanation / Answer
velocity of m1 before making collision with m2,
u1 = sqrt(2*g*h)
velocity of m2 before making collision with m1,
u2 = -sqrt(2*g*(h - h/2))
= -sqrt(g*h)
so, u1 = -sqrt(2)*u2
v1f = (m1 - m2)*u1/(m1+m2) + 2*m2*u2/(m1+m2)
= (m1 - 9*m1)*u1/(m1+9*m1) + 2*9*m1*(-u1/sqrt(2))/(m1+9*m1)
= -(0.8 + 1.8/sqrt(2) )*u1
= -(0.8 + 1.8/sqrt(2) )*sqrt(2*g*h)
= -1.93*sqrt(g*h)
v2f = (m2-m1)*u2/(m1+m2) + 2*m1*u1/(m1+m2)
= (9*m1 - m1)*u2/(m1+9*m1) + 2*m1*(-sqrt(2)*u2)/(m1+9*m1)
= 0.8*u2 - 0.2*sqrt(2)*u2
= 0.8*(-sqrt(g*h)) - 0.2*sqrt(2)*(-sqrt(g*h))
= -0.517*sqrt(g*h)