A 3 kg chunk of putty moving at a speed of 5 m/s collides with and sticks to a 5 kg bowling ball that is initially at rest. What is their final velocity?

Please explain. I am really confused. I know you have to use the conservation of momentum equation:

\[ v_f = \frac{m_1 \cdot v_1}{m_1 + m_2} \]

where:
- \( m_1 \) is the mass of the putty,
- \( v_1 \) is the velocity of the putty,
- \( m_2 \) is the mass of the bowling ball,
- \( v_f \) is the final velocity after collision.

Please help!

You'd have an easier time using the equation if you understood where the equation comes from.
The law here ... the major principle to remember, the key, the fundamental truth, the big cookie ... is the fact that momentum is conserved . The total momentum after they join up is the same as the total momentum before they meet.
Momentum of an object is (mass) times (speed).
Now, list all the things you know, before and after the putty meets the ball:
Before: ** **There are two objects. Mass of putty = 3 kg Speed of putty = 5m/s Momentum of putty = 3 x 5 = 15 kg-m/s.
Mass of ball = 5 kg Speed of ball = zero Momentum of ball = 5 x 0 = zero
Total momentum of both things = 15 kg-m/s
After : There is only one object, because they stuck together. Mass of (putty+ball) = (3+5) = 8 kg Speed of (putty+ball) = we don't know; that's what we have to find
Momentum of (putty+ball) = 8 x (speed)
===================================
We know that the momentum after is equal to the momentum before.
8 x (speed) = 15 kg-m/s
Divide each side by 8 :
Speed = 15 / 8 = ***1.875 m/s *** after they stick together.

Answered by AL2006 | 2024-06-10

The final velocity of the putty and bowling ball after they collide and stick together is 1.875 m/s. This is determined using the conservation of momentum, where the total momentum before the collision is equal to the total momentum after the collision. The calculation shows that they move together with a common velocity of 1.875 m/s after the collision.
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Answered by AL2006 | 2024-12-26