A bullet is fired through a wooden board with a thickness of 9.0 cm. The bullet hits the board perpendicular to it, and with a s
1 answer:
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We can solve the problem by using conservation of momentum.
The player + ball system is an isolated system (there is no net force on it), therefore the total momentum must be conserved. Assuming the player is initially at rest with the ball, the total initial momentum is zero:
The total final momentum is:
where is the momentum of the player and
is the momentum of the ball.
The momentum of the ball is:
While the momentum of the player is: , where M=59 kg is the player's mass and vp is his velocity. Since momentum must be conserved,
so we can write
and we find
and the negative sign means that it is in the opposite direction of the ball.
Answer:
+17 kg m/s
Explanation:
Question is missing. Found it on google:
"<em>What is the magnitude of the final momentum of the bowling pin if it has a mass of 1.5 kg</em>?"
Solution:
we can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the system must be conserved, so we can write:
where
is the momentum of the ball before the collision
is the momentum of the pin before the collision (zero because the pin is stationary)
is the momentum of the ball after the collision
is the momentum of the pin after the collision
Solving the equation for , we find: