A 0.400 kg block sliding on a frictionless floor with a velocity of 3.00 m/s Eas
ID: 1546558 • Letter: A
Question
A 0.400 kg block sliding on a frictionless floor with a velocity of 3.00 m/s East has an elastic collision with a 0.600 kg block that has a velocity of 5.50 m/s West. The 0.400 kg block then hits a spring with a spring constant of 332 N/m, slowing to a stop, while the 0.600 kg block has a completely inelastic collision with a 0.800 kg black that has a velocity of 2.20 m/s West. a.) What are the magnitudes and directions of the velocities of the 0.400 kg block and the 0.600 kg block right after the elastic collision? b.) By how many meters is the spring compressed? c) What is the magnitude and direction of the velocity of the 0.600 kg block and 0.800 kg block after the completely inelastic collision?Explanation / Answer
by conservation of momentum
initial momentum = final momentum
0.4 * 3 - 0.6 * 5.5 = 0.4 * v1 + 0.6 * v2 --------(1)
by conservation of energy
initial energy = final energy
0.5 * 0.4 * 3^2 + 0.5 * 0.6 * 5.5^2 = 0.5 * 0.4 * v1^2 + 0.5 * 0.6 * v2^2 -----------(2)
on solving 1 and 2 we'll get
v1 = -7.2 m/s
v2 = 1.3 m/s
a) velocity of 0.4 kg block = 7.2 m/s toward west
a) velocity of 0.6 kg block = 1.3 m/s toward east
by conservation of energy
0.5 * kx^2 = 0.5 * mv^2
0.5 * 332 * x^2 = 0.5 * 0.4 * 7.2^2
x = 0.2499 m
b) spring is compressed by 0.2499 m
by conservation of energy
initial energy = final energy
0.6 * 1.3 = (0.4 * 0.6) * v
v = 0.78 m/s
c) velocity of blocks after collision = 0.78 m/s toward east
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