(a) If the block moves 4.80 cm to the right after impact, find the speedat which
ID: 1753741 • Letter: #
Question
(a) If the block moves 4.80 cm to the right after impact, find the speedat which the bullet emerges from the block.m/s
(b) If the block moves 4.80 cm to theright after impact, find the mechanical energy converted intointernal energy in the collision.
2 J A 5.00 g bullet moving with an initial speed of 370 m/s is fired into and passes through a 1.00 kgblock. The block, initially at rest on a frictionless, horizontalsurface, is connected to a spring of force constant 915 N/m. (a) If the block moves 4.80 cm to the right after impact, find the speedat which the bullet emerges from the block. m/s (b) If the block moves 4.80 cm to theright after impact, find the mechanical energy converted intointernal energy in the collision. 2 J
Explanation / Answer
a)
from the given problem first we find the speed whenthe bullet emerges from the block by using
the law of conservation of momentum
m vi = M Vi + m v
from the given the block moves a distance 5.00 cm, andthe block quickly reaches the maximum
velocity Vi and the bullet will be goingwith constant velocity v, the block then compresses the
spring and stops so we get
(1 / 2) M Vi2 = (1 / 2) kx2
Vi = [(915 N / m) (0.048m)2 / (1.00 kg)]
= ....... m /s
the bullet emerges from the block at a speed of
v = (m vi - M Vi) / m
= [(0.005 kg) (370 m / s) - (1.00kg) (Vi)] / (0.005 kg)
= ....... m / s
(b)
according to the law of conservation of energy weget
E = KE + PE
= (1 / 2) (0.005kg) (v2) - (1 / 2) (0.005 kg) (370 m / s) + (1 / 2) (915N / m) (0.048 m)2
= ....... J
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