A lead ball of mass 60kg hangs from a string. A piece of putty mass of 15kg hits the ball with velocity of 12m/s. If the putty sticks to the ball, what will be their velocity immediately after the impact?
M1*V1 + M2*V2 = M1*V3 + M2*V3
V3 = V4.
60*0 + 15*12 = 60*V3 + 15*V3
180 = 75V3
V3 = 2.4 m/s = Velocity after impact.
To determine the velocity of the lead ball and the piece of putty immediately after the impact, we can use 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.
The momentum of an object is calculated by multiplying its mass by its velocity:
Momentum = mass × velocity
Before the collision, the lead ball is hanging and not moving, so its momentum is zero. However, the piece of putty is moving with a velocity of 12 m/s and has a mass of 15 kg. Therefore, the momentum of the putty before the collision is:
Momentum of putty before = mass of putty × velocity of putty
= 15 kg × 12 m/s
= 180 kg·m/s
Since the putty sticks to the lead ball, the two objects become a single system. Let's assume the velocity immediately after the impact is "v" (which is the final velocity of both the lead ball and the putty together).
After the collision, the combined mass of the lead ball and the putty is:
Total mass after = mass of lead ball + mass of putty
= 60 kg + 15 kg
= 75 kg
Now, using the principle of conservation of momentum, we can equate the total momentum before the collision to the total momentum after the collision:
Total momentum before = Total momentum after
0 + 180 kg·m/s = 75 kg × v
Solving for "v", we find:
v = (0 + 180 kg·m/s) ÷ 75 kg
= 2.4 m/s
Therefore, immediately after the impact, the combined system of the lead ball and the piece of putty will have a velocity of 2.4 m/s.