1. A 4.0 kg block moving with a velocity of +4.0 m/s makes an elastic collision
ID: 1597152 • Letter: 1
Question
1. A 4.0 kg block moving with a velocity of +4.0 m/s makes an elastic collision with a stationary block of mass 2.1 kg.
(a) Use conservation of momentum and the fact that the relative speed of recession equals the relative speed of approach to find the velocity of each block after the collision.
-----m/s (for the 4.0 kg block)
-----m/s (for the 2.1 kg block)
(b) Check your answer by calculating the initial and final kinetic energies of each block.
----- J (initially for the 4.0 kg block)
----- J (initially for the 2.1 kg block)
----- J (finally for the 4.0 kg block)
----- J (finally for the 2.1 kg block)
2. A 3.1 kg block moving with a velocity of +3.6 m/s makes an elastic collision with a stationary block of mass 1.6 kg.
(a) Use conservation of momentum and the fact that the relative speed of recession equals the relative speed of approach to find the velocity of each block after the collision.
------m/s (for the 3.1 kg block)
-----m/s (for the 1.6 kg block)
(b) Check your answer by calculating the initial and final kinetic energies of each block.
------J (initially for the 3.1 kg block)
------J (initially for the 1.6 kg block)
-----J (finally for the 3.1 kg block)
-----J (finally for the 1.6 kg block)
Both of these are the same with different numbers. Thanks in advance!
Explanation / Answer
1) a) using conservation of momentum
u1-u2 = v2-v1
v1 = 2m2u2+(m1-m2)u1/(m1+m2)
if u2 = 0
v1 = (4-2.1)4/6.1 = 1.246 m/s
v2 = 2m1u1+(m2-m1)u2/(m1+m2)
if u2 = 0
v2 = 2*4*4/6.1 = 5.246 m/s
k1i = 1/2*4*16 = 32 J
k2i = 0
ki = 32 J
k1f = 3.11 J
k2f = 28.89 J
kf = 32 J
ki = kf
2) a) v1 = (m1-m2)u1/(m1+m2) = 1.149 m/s
v2 = 2m1u1/(m1+m2) = 4.749 m/s
k1i =20.1 J
k2i = 0
ki = 20.1 J
K1I = 2.05 J
k2f = 18.04 J
ki = kf
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