Initially, ball 1 rests on an incline of height h, and ball 2 rests on an inclin
ID: 1640574 • 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 = 5m_1, m_1 = 0.046 kg, and h = 0.77 m, what is the velocity of each ball after the collision? (Assume the balls slide but do not roll. Indicate e direction with the sign of your answer. Positive is to the right, and negative 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 vector _1f = m/s v vector _2f = m/sExplanation / Answer
First of all we will calculate the velocity of the two ball by energy conservation .
Initial energy of the ball = potential energy = mgh
final energy of the ball = kinetic energy = (1/2)mu2
equating the initial and final energy
u = (2gh)1/2 ,
hence the velocity of the ball 1 before the collision u1 = (2gh)1/2 = 3.887 m/s
Velocity of the ball 2 before the collision u2 = (2g(h/2))1/2 = 2.748 m/s
Now the direction of the velocity of the ball 1 is toward +X therefore (u1) = 3.887 m/s
But the direction of the ball 2 is toward -X , therefore (u2) = -2.748 m/s
Now we know that the momentum remain in conserved in the collision , therefore
Initial momentum = m1u1 + m2u2 = m1u1 + 5m1u2 ------------(1)
Final momentum = m1V1 + m2V2 = m1V1 + 5m1V2 --------------(2)
Initially we consider that the ball moves in the + X direction after the collision ,
and V represent the velocity after the collision
Equating 1 and 2
m1V1 + 5m1V2 = m1u1 + 5m1u2
V1 + 5V2 = u1 + 5u2 = 3.887 - (5*2.748) = -9.853 ------------(3)
We also know that in the elastic collision
velocity of approach = velocity of seperation
u1 - u2 = V2 - V1
V2 - V1 = 3.887 -(-2.748) = 6.635 -------------(4)
solving 3 and 4
V2 = -0.536 m/s
V1 = -7.171 m/s