A 3.1 kg block moving with a velocity of +3.5 m/s makes an elastic collision wit

Question

A 3.1 kg block moving with a velocity of +3.5 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)

Are the two total kinetic energies the same? Yes No

Explanation / Answer

m1 = 3.1 kg
u1 = +3.5 m/s

m2 = 1.6 kg
u2 = 0.0 m/s

(A) Elastic collision with stationary object, solving for the first object (3.5 kg):

v1 = { [m1-m2] / [m1+m2] } * u1

v1 = { [ (3.1 kg) - (1.6 kg) ] / [ (3.1 kg) + (1.6 kg) ] } * (+3.5 m/s)
v1 = { [ 1.5 kg ] / [ 4.7 kg ] } * (+3.5 m/s)
v1 = { 0.319 } * (+3.5 m/s)
v1 = 1.12 m/s

for block of mass 1.6kg

v2 = { [2m1] / [m1+m2] } * u1
v2 = { [2 * (3.1 kg) ] / [ (3.1 kg) + (1.6 kg) ] } * (+3.5 m/s)
v2 = { [ 6.2 kg ] / [ 4.7 kg ] } * (+3.5 m/s)
v2 = { 1.319 } * (+3.5 m/s)
v2 = 4.62 m/s

(B) Kinetic energies as given by KE = m * v^2

3.1kg block pre-impact
KE = 0.5 * (3.1 kg) * (3.5 m/s)^2
KE = (1.55 kg) * (12.25 m^2/s^2)
KE = 18.99 J

1.6 kg block pre-impact
KE = 0.5 * (1.6 kg) * (0.0 m/s)^2
KE= 0 J

3.5 kg block post-impact
KE = 0.5 * (3.1 kg) * (1.12 m/s)^2
KE = (1.55 kg) * (1.254 m^2/s^2)
KE = 1.94 J

1.6 kg post impact
KE = 0.5 * (1.6 kg) * (4.62 m/s)^2
KE = (0.8 kg) * (21.34 m^2/s^2)
KE = 17.08 J

Pre impact,
(18.99 J) + (0.0 J) = 18.99 J =19J

Post impact,
(1.94 J) + (17.08 J) = 19.01 J=19J

the two total kinetic energies are same.

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