A metal block of mass A/= 1.99 kg resting on a horizontally as shown below is hi
ID: 1555253 • Letter: A
Question
A metal block of mass A/= 1.99 kg resting on a horizontally as shown below is hit by a bullet of mass m = 1.00 times 10^-2 kg shot vertically from underneath. At the moment of impact the bullet was moving upward with a speed v = 350 m/s. After the impact, the block jumps to height of 0.625 m above the surface. a) Find the velocity of the block acquired during the impact. b) Find velocity of the bullet after the impact. c) Determine the type of collision based on your findings (elastic vs inelastic). The impact happens instantaneous, and effects of air resistance obviously should not be considered. Acceleration of gravity g = 9.80 m/s^2.Explanation / Answer
after collision the block rised to height h
kinetic energy of the block after collision = potential energy of the block at h
(1/2)*M*Vblock^2 = M*g*h
(1/)*Vblock^2 = g*h
Vblock = 3.5 m/s
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(b)
from momentum conservation
momentum before collision = momentum after collision
m*v = m*vbullet + M*vblock
10^-2*350 = (10^-2*Vbullet) + (1.99*3.5)
vbullet = -346.5 m/s
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initial kinetic energy Ki = (1/2)*mbullet*v^2 = (1/2)*10^-2*350^2 = 612.5 J
final KE = Kf = (1/2)*mbullet*vbullet^2 + (1/2)*M*vblock^2
Kf = ((1/2)*10^-2*346.5^2) + ((1/2)*1.99*3.5^2)
Kf = 612.5 J
Kf = Ki
collison is ELASTIC