Fig. 2.1 shows a snooker ball of mass 0.40 kg travelling at 2.0 m/s colliding head on with another identical ball travelling at 1.0 m/s. After the collision, the first snooker ball bounces back at 0.50 m/s. 10 snooker ball 2.0 m/s snooker ball 101 1.0 m/s Fig. 2.1 (1) Show that the second ball bounces back as well.
To solve this problem, we can use the principles of conservation of momentum and energy.
Let's define the following variables:
m1 = mass of the first snooker ball (0.40 kg)
v1i = initial velocity of the first snooker ball (2.0 m/s)
v1f = final velocity of the first snooker ball (0.50 m/s)
m2 = mass of the second snooker ball (0.40 kg)
v2i = initial velocity of the second snooker ball (1.0 m/s)
v2f = final velocity of the second snooker ball (unknown)
According to the conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision:
m1 * v1i + m2 * v2i = m1 * v1f + m2 * v2f
Substituting the given values:
(0.40 kg)(2.0 m/s) + (0.40 kg)(1.0 m/s) = (0.40 kg)(0.50 m/s) + (0.40 kg)(v2f)
0.80 kg m/s + 0.40 kg m/s = 0.20 kg m/s + 0.40 kg * v2f
1.20 kg m/s = 0.20 kg m/s + 0.40 kg * v2f
0.40 kg * v2f = 1.00 kg m/s
v2f = 1.00 kg m/s / 0.40 kg
v2f = 2.50 m/s
Therefore, the second snooker ball bounces back with a velocity of 2.50 m/s.
To show that the second ball bounces back as well, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision should be equal to the total momentum after the collision.
Let's consider the momentum of the system before the collision. The total momentum is given by the sum of the individual momentum of each ball.
Before the collision:
- Mass of each ball (m) = 0.40 kg (given)
- Velocity of ball 1 (v1) = 2.0 m/s (given)
- Velocity of ball 2 (v2) = 1.0 m/s (given)
Momentum of ball 1 before the collision = m * v1 = 0.40 kg * 2.0 m/s = 0.80 kg·m/s
Momentum of ball 2 before the collision = m * v2 = 0.40 kg * 1.0 m/s = 0.40 kg·m/s
Total momentum before the collision = momentum of ball 1 + momentum of ball 2 = 0.80 kg·m/s + 0.40 kg·m/s = 1.20 kg·m/s
Now, let's consider the momentum after the collision.
- Velocity of ball 1 after the collision (v'1) = -0.50 m/s (given, because it is bouncing back with a negative velocity)
- Velocity of ball 2 after the collision (v'2) = ? (to be determined)
Using the conservation of momentum principle, the total momentum after the collision should be equal to the total momentum before the collision.
Total momentum after the collision = momentum of ball 1 after the collision + momentum of ball 2 after the collision
0 = m * v'1 + m * v'2
0 = 0.40 kg * (-0.50 m/s) + 0.40 kg·m/s
Now, we can solve for the velocity of ball 2 after the collision (v'2).
0 = -0.20 kg·m/s + 0.40 kg·m/s + 0.40 kg·m/s
0 = 0.60 kg·m/s
Therefore, the second ball bounces back with a velocity of 0.60 m/s.
Since the velocity of the second ball after the collision is not zero, it indicates that the second ball bounces back as well.