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a 0.030-kg bullet is fired vertically at 200 m/s into a 0.15-kg baseball that is initially at rest. how high does the combined bullet and baseball rise after the collision, assuming the bullet embeds itself in the ball?

Sagot :

If a 0.030-kg bullet is shot vertically at 200 m/s into a 0.15-kg baseball, the combined height of the bullet and baseball following the collision is at most 56.6 m, provided the bullet embeds itself in the ball.

This phase will begin immediately prior to the collision and conclude immediately following the impact (when the two objects are on the verge to move up as a single unit). Note that we are assuming the ball will initially be moving at a speed of 200 m/s when the bullet then collision with it. At this point, we define the system as the ball plus the bullet. We are ignoring any outside factors, such as air resistance or even gravitational force, that can affect this system. As a result, the final velocity of the system immediately following the impact can be determined using the conservation of momentum principle, p=0. The gravitational potential energy is being ignored because we assume that the ball and the bullet were at the same level just before (and just after) colliding.

Yf = v1^2*2*9.8

Yf = 33.3^2*2*9.8

Yf = 56.6 metres.

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