HELP!!!A pitcher throws a 0.5kg ball of clay at a 6kg block of wood. The clay sticks to the wood on impact and their joint velocity afterwards is 3m/s. What is the original speed of the clay? I know the answer is 39m/s but I have no idea how they come up with that
To find the original speed of the clay before 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.
Let's denote:
- m1 as the mass of the clay (0.5 kg),
- v1 as the original velocity of the clay (which we're trying to find),
- m2 as the mass of the wooden block (6 kg),
- v2 as the original velocity of the wooden block (assuming it's initially at rest),
- vf as the final velocity of both the clay and the block after the collision (3 m/s).
Using the conservation of momentum equation:
m1 * v1 + m2 * v2 = (m1 + m2) * vf
Plugging in the known values:
0.5 kg * v1 + 6 kg * 0 = (0.5 kg + 6 kg) * 3 m/s
0.5 kg * v1 = 6.5 kg * 3 m/s
0.5 kg * v1 = 19.5 kg * m/s
Now we can solve for v1 by dividing both sides of the equation by 0.5 kg:
v1 = (19.5 kg * m/s) / 0.5 kg
v1 = 39 m/s
Therefore, the original speed of the clay before the collision is 39 m/s.
conservation of momentum
initialmomentum=final
.5*V=(.5+6)3
solve for V