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A block of wood mass 0.60 kg is balanced on top of a vertical port 2.0 m high. A 10 gm
bullet is fired horizontally into the block and the embedded bullet land at a 4.0 m from
the base of the port. Find the initial velocity of the bullet.
A. 10 3 m/s B. 8.32 m/s C. 3.82 m/s D. 1.0 m/s

Respuesta :

Answer:

m V1 = (m + M) V2 conservation of momentum

T = time to fall 2 m

2 = 9.8 / 2 T^2    initial vertical velocity zero

T^2 = 4/9.8       T = .639 sec

Sx = V2 * .639 sec     distance traveled horizontally

V2 = 4 /.639 = 6.26 m/sec

V1 = (m + M) / m * 6.26 m/sec       from first equation

V1 = (600 + 10) / 10 * 6.26 m/s = 382 m/s

There seems to be some confusion on units

Note that 382 m/s is a reasonable answer for speed of bullet

1100 ft/sec is reasonable for a .22 cal bullet

1100 ft/sec / 3.28 ft/m = 335 m/sec

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