At a wild-west show, a marksman fires a bullet at a 12 g coin that\'s thrown str

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 550 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 150 m/s . At what angle measured with respect to the horizontal does the bullet ricochet away from this collision?

Can you please give me the step-by-step solution for this problem..Thanks.

Explanation / Answer

At highest point of coin, Velocity of coin is 0.
Initial Velocity of bullet, Vx = 550 * cos(45)
Initial Velocity of bullet, Vy = 550 * sin(45)  

Let the Velocity of Bullet be Vx i^ + Vy j^
So Using Momentum Conservation,
Initial Momentum = Final Momentum

In X direction,
15* 550 * sin(45) = 15 * v2x
v2x =  388.9 m/s
vxf * cos() = 388.9 m/s ---------1

In Y direction,
15* 550 * sin(45) = 15 * v2y + 12 * 150
15 * v2y = 4033.6
vxf * sin() = 268.9 m/s ----------2

tan() = 268.9/388.9
= 34.66 o Above Horizontal.

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