A 0.2kg bullet travelling at 250m/s goes right through a 10kg block of wood emer

Question

A 0.2kg bullet travelling at 250m/s goes right through a 10kg block of wood emerging with a speed of 50 m/s. In the first instance, the block is suspended by a vertical string of length 8.82 m. attached to a hook in the ceiling and is free to rotate like a pendulum. With what initial speed does the block move after the bullet has passed through it? To what height doe it rise before stopping and then swinging like a pendulum? What angle does the string make with the vertical direction, when the block is at its highest point? Now suppose the block is sitting on a horizontal surface (No longer attached to the string) so that after the bullet emerges, the block slides along the surface, with coefficients of friction 0.3 and 0.35 respectively. How far does the block slide before coming to rest?

Explanation / Answer

apply the law of conservation of moemntun as

m1u1 + m2u2 = (m1 + m2) V

so

V = (0.2 * 250) + ( 10* 50)/(10+0.2)

V = 53.92 m/s

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apply Hmax = u^2/2g

Hmax = 53.92 * 53.92/(2*9.8)

Hmax = 148.33 m

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apply tna theta =   53.92/(250)

tan theta = 0.2158

theta = 12.17 deg

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