Question

At a wild-west show, a marksman fires a bullet at a 12 g coin that's thrown straight up into the air. The marksman points his rifle at a 45 ∘ angle above the ground, then fires a 15 g bullet at a speed of 510 m/s. Just as the coin reaches its highest point, the bullet hits it and glances off, giving the coin an exactly vertical upward velocity of 100 m/s.

At what angle measured with respect to the horizontal does the bullet ricochet away from this collision?

Answer 1

The given setup is like this:

I will assume that the bullet hits the coin with a negligible time elapsed, to ignore the affect of gravity on the bullet.

To make things simpler, take a different frame of reference like this:

The bullet hits the coin when the coin is at the highest point. Any object at it's highest point will have zero vertical velocity. It has no horizontal velocity either.

The elastic collision equation required is:

m1=15 g, m2 = 12 g

In the x direction: (i.e. along path of bullet)

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v2 is obtained from the fact that the coin after getting hit by the bullet acquires 100m/s purely in the vertical direction in the original frame. Since this new frame is tilted by 45 degrees, the final velocity of the coin is 45 degree with respect to the horizontal here.

Thus, the vertical and horizontal components of v2 are each equal to

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Thus,coming back to the equation, we have:

In the y direction:

we get:

Thus, the angle is given by:

This is the angle in the rotated frame.

The angle in the original frame is thus:

angle = 45 - 7.111 = 37.88 degrees

Thus the bullet will finally travel at 37.88 degrees w.r.t the horizontal