Initially, ball 1 rests on an incline of height h, and ball 2 rests on an inclin
ID: 1419196 • 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 elastically in the trough of the track. If m_2 = 9 m_1, m_1 = 0.041 kg, and h = 0.78 m, what is the velocity of each ball after the collision? (Assume the balls slide but do not roll. Indicate the direction with the sign of your answer. Positive is to the right, and negative is to the left. Due to the nature of this problem, do not use rounded intermediate values in your calculations-including answers submitted in WebAssign.) v_1f = m/s v_2f = m/sExplanation / Answer
m2 =9*m1 = 0.369 kg
before collision
conservation energy theorem
mgh = 1/2mv^2
u1 = sqrt(2gh)
u1 =sqrt ( 2*9.8*0.78 ) = 3.91 m/s
u2 = sqrt(2gh/2) = sqrt ( 9.8*0.78 )
u2 = - 2.765 m/s
after collision
v1f = [((m1 - m2 )*u1) + ( 2m2*u2) ] / (m1 + m2)
=[((0.041 - 0.369 )*3.91) + ( 2*0.369* (- 2.765)) ] / (0.041 + 0.369)
v1f = - 8.105 m/s
v2f = [(2m1u1) + (m2-m1)*u2] / (m1 + m2)
v2f = [(2*0.041*3.91) + (0.369-0.041)*(- 2.765)] / (0.041 + 0.369)
v2f = -1.43 m/s