A billiard ball is given an initial velocity V=5M/S
and makes a direct impact with ball B (e=0.8), calculate the velocity of ball B and the rebound angle after it hits cushion at C (e =0.6). Mass of the balls is 0.4 kg.
The Correct Answer and Explanation is:
To solve this, we can break it into two parts: the collision between the billiard balls and the ball’s rebound from the cushion.
Part 1: Collision between Ball A and Ball B
Given:
- Velocity of ball A before the collision, VA=5 m/sV_A = 5 \, \text{m/s}VA=5m/s
- Coefficient of restitution between the balls, e=0.8e = 0.8e=0.8
- Mass of the balls, mA=mB=0.4 kgm_A = m_B = 0.4 \, \text{kg}mA=mB=0.4kg
- Assume a one-dimensional collision for simplicity (in a straight line).
The coefficient of restitution eee is related to the velocities of the balls before and after the collision by the equation:e=VB′−VA′VA−VBe = \frac{V_{B}’ – V_{A}’}{V_A – V_B}e=VA−VBVB′−VA′
Where:
- VAV_AVA is the velocity of ball A before the collision
- VBV_BVB is the velocity of ball B before the collision
- VA′V_A’VA′ and VB′V_B’VB′ are the velocities of balls A and B after the collision
Assuming ball B is initially at rest ( VB=0V_B = 0VB=0 ):e=VB′−VA′VAe = \frac{V_{B}’ – V_A’}{V_A}e=VAVB′−VA′
Since we have only one equation and two unknowns, we also need to use the conservation of momentum:mAVA+mBVB=mAVA′+mBVB′m_A V_A + m_B V_B = m_A V_A’ + m_B V_B’mAVA+mBVB=mAVA′+mBVB′
Substituting mA=mBm_A = m_BmA=mB and VB=0V_B = 0VB=0, the equation becomes:0.4⋅5+0.4⋅0=0.4⋅VA′+0.4⋅VB′0.4 \cdot 5 + 0.4 \cdot 0 = 0.4 \cdot V_A’ + 0.4 \cdot V_B’0.4⋅5+0.4⋅0=0.4⋅VA′+0.4⋅VB′2=0.4⋅VA′+0.4⋅VB′2 = 0.4 \cdot V_A’ + 0.4 \cdot V_B’2=0.4⋅VA′+0.4⋅VB′
This simplifies to:5=VA′+VB′5 = V_A’ + V_B’5=VA′+VB′
Now, we have two equations:
- e=VB′−VA′5=0.8e = \frac{V_B’ – V_A’}{5} = 0.8e=5VB′−VA′=0.8
- 5=VA′+VB′5 = V_A’ + V_B’5=VA′+VB′
From the first equation:VB′−VA′=4V_B’ – V_A’ = 4VB′−VA′=4
Now we solve these equations simultaneously:
- VA′+VB′=5V_A’ + V_B’ = 5VA′+VB′=5
- VB′−VA′=4V_B’ – V_A’ = 4VB′−VA′=4
Adding the two equations:2VB′=92 V_B’ = 92VB′=9VB′=4.5 m/sV_B’ = 4.5 \, \text{m/s}VB′=4.5m/s
Substituting into VA′+VB′=5V_A’ + V_B’ = 5VA′+VB′=5:VA′+4.5=5V_A’ + 4.5 = 5VA′+4.5=5VA′=0.5 m/sV_A’ = 0.5 \, \text{m/s}VA′=0.5m/s
Part 2: Rebound from the Cushion
After ball B hits the cushion at point C, the velocity component perpendicular to the cushion will reverse, while the component parallel to the cushion will remain unchanged. The coefficient of restitution between ball B and the cushion is e=0.6e = 0.6e=0.6.
If the angle of incidence (the angle between the velocity vector and the cushion’s surface) is θ\thetaθ, then the rebound angle will be the same as the angle of incidence, but in the opposite direction.
The velocity of ball B after the rebound in the direction perpendicular to the cushion is given by:V⊥′=e⋅V⊥V_{\perp}’ = e \cdot V_{\perp}V⊥′=e⋅V⊥
Where V⊥V_{\perp}V⊥ is the component of the velocity perpendicular to the cushion.
Since the parallel component remains unchanged, the rebound velocity components are:
- Parallel component: V∥′=V∥V_{\parallel}’ = V_{\parallel}V∥′=V∥
- Perpendicular component: V⊥′=0.6⋅V⊥V_{\perp}’ = 0.6 \cdot V_{\perp}V⊥′=0.6⋅V⊥
Thus, the velocity of ball B after the rebound will have the same magnitude but will change direction due to the reversal of the perpendicular component.
Final Velocity and Rebound Angle
- The velocity of ball B after the collision is VB′=4.5 m/sV_B’ = 4.5 \, \text{m/s}VB′=4.5m/s
- The rebound angle after hitting the cushion will depend on the angle of incidence, which is not provided. If the angle of incidence is θ\thetaθ, the rebound angle will also be θ\thetaθ but in the opposite direction.
