9Grant's basketball has a mass 0.560 kg and rolls along the ground at 7.6 m/s towards a stationary bowling ball of mass 4.5 kg. After the collision, the bowling ball travels at 1.5 m/s. What is the velocity of the Grant's basketball?
To solve this problem, we can use the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision.
Let the initial velocity of the Grant's basketball be V1 and the final velocity of Grant's basketball be V2.
Initial momentum before the collision:
m1 * V1 + m2 * 0 = (m1 + m2) * V_final
0.560 kg * 7.6 m/s + 4.5 kg * 0 = (0.560 kg + 4.5 kg) * 1.5 m/s
4.256 + 0 = 5.06 * 1.5
4.256 = 7.59
Therefore, the velocity of Grant's basketball after the collision is 7.59 m/s.