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.
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.
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