A stationary billiard ball, mass 0.17 kg, is struck by an identical ball moving at 2.7 m/s. After the collision, the second ball moves off at 60° to the left of its original direction. The stationary ball moves off at 30° to the right of the moving ball's original direction. What is the speed of each ball after the collision?
You have three equations
momentum to the right/left
momentum up/down
conservation of energy
Two unknowns: Va and Vb.
on the momemtum up/down, write it as a function of Va and Vb times sinAngle
one the momentum right/left, use Va and Vb times Cosangle.
It will solve, the algebra is messy at times
I'm sorry, but that doesn't make sense to me, is there any other way you can explain it?
Thanks.
They are asking for only two unknowns although three equations could be applied. You can solve the problem just by using momentum conservation, but presumably kinetic energy will also be conserved, if the angles they provided are correct. (Billiard ball collisions are very elastic)
M Vo = M Va cos 30 + M Vb cos 60
0 = M Va sin 30 - M Vb sin 30
Vo = 0.866 Va + 0.500 Vb
0.500 Va = 0.866 Vb
Vo = 0.866 Va+(0.250/0.866)Va
= 1.1546 Va
Va = 0.866 Vo
Vb = 0.500/0.866 * 0.866 Vo = Vo/2
Note that Va^2 + Vb^2 = Vo^2, so energy is indeed conserved.
To solve this problem, we will make use of the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
Let's calculate the total momentum before the collision of the two balls.
The momentum of an object is calculated by multiplying its mass by its velocity.
Momentum = mass × velocity
The first ball is stationary, so its velocity is zero. Therefore, its momentum is zero.
The second ball has a mass of 0.17 kg and a velocity of 2.7 m/s. Thus, its momentum is:
Momentum of second ball = 0.17 kg × 2.7 m/s = 0.459 kg·m/s
Now, let's find the total momentum after the collision.
After the collision, the second ball moves off at 60° to the left of its original direction. The stationary ball moves off at 30° to the right of the moving ball's original direction.
To calculate the momentum of each ball after the collision, we need to break down their velocities into horizontal and vertical components.
The horizontal component of the second ball's velocity is calculated as:
Horizontal component = velocity × cos(angle)
Horizontal component of the second ball's velocity = 2.7 m/s × cos(60°) = 2.7 m/s × 0.5 = 1.35 m/s
The vertical component of the second ball's velocity is calculated as:
Vertical component = velocity × sin(angle)
Vertical component of the second ball's velocity = 2.7 m/s × sin(60°) = 2.7 m/s × sqrt(3)/2 = 2.338 m/s
Similarly, we can calculate the horizontal and vertical components of the velocity of the stationary ball after the collision.
The horizontal component of the stationary ball's velocity is calculated as:
Horizontal component = velocity × cos(angle)
Horizontal component of the stationary ball's velocity = velocity × cos(30°) = velocity × 0.866
The vertical component of the stationary ball's velocity is calculated as:
Vertical component = velocity × sin(angle)
Vertical component of the stationary ball's velocity = velocity × sin(30°) = velocity × 0.5
Since the two balls move off in opposite directions, the total momentum after the collision is the difference between their individual momenta.
Total momentum after collision = Momentum of second ball - Momentum of stationary ball
Total momentum after collision = 0.459 kg·m/s - (-0.17 kg) × (velocity × 0.866 - velocity × 0) - velocity × 0.5)
To find the velocity of each ball after the collision, we can set up the equation:
Total momentum before collision = Total momentum after collision
0 kg·m/s = 0.459 kg·m/s - 0.17 kg·m/s - velocity × 0.5)
Rearranging the equation, we get:
velocity × 0.5 = 0.459 kg·m/s - 0.17 kg·m/s
velocity × 0.5 = 0.289 kg·m/s
Dividing both sides of the equation by 0.5, we get:
velocity = 0.289 kg·m/s ÷ 0.5
velocity = 0.578 m/s
Therefore, the speed of each ball after the collision is approximately 0.578 m/s.