A 3.0 kg block moving with a velocity of +4.8 m/s makes an elastic collision wit

Question

A 3.0 kg block moving with a velocity of +4.8 m/s makes an elastic collision with a stationary block of mass 2.4 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 3.0 kg block) m/s (for the 2.4 kg block) (b) Check your answer by calculating the initial and final kinetic energies of each block. J (initially for the 3.0 kg block) J (initially for the 2.4 kg block) J (finally for the 3.0 kg block) J (finally for the 2.4 kg block)

Explanation / Answer

Given values:

m1 = 3 kg
u1 = 4.8 m/s

m2 = 2.4 kg
u2 = 0.0 m/s

(A) Post-impact velocities

v'1 = { [ m1 - m2 ] / [ m1 + m2 ] } * u1

v'1 = { [ (3kg) - (2.4 kg) ] / [ (3 kg) + (2.4 kg) ] } * (4.8 m/s)
v'1 = { [ 0.6 kg ] / [ 5.4 kg ] } * (4.8 m/s)
v'1 = { 0.11 } * (4.8 m/s)
v'1 = 0.53 m/s

v'2 = { [ 2 * m1 ] / [ m1 + m2 } } * u1

v'2 = { [ 2 * (3kg) ] / [ (3 kg) + (2.4 kg) ] } * (4.8 m/s)
v'2 = { [ 6 kg ] / [ 5.4 kg ] } * (4.8 m/s)
v'2 = { 1.11} * (4.8 m/s)
v'2 = 5.33 m/s

(B) Kinetic energies, using KE = 0.5 * m * v^2

KE Block #1 Initial = 0.5 * (3 kg) * (4.8 m/s)^2
KE Block #1 Initial = (1.5 kg) * (23.04 m^2/s^2)
KE Block #1 Initial = 34.56 J

KE Block #1 Initial = 0 J, because it wasn't moving

Total energy of initial system = 34.56 J + 0 J = 34.56 J

KE Block #1 Final = 0.5 * (3kg) * (0.53 m/s)^2
KE Block #1 Final = (1.5 kg) * ( 0.281 m^2/s^2)
KE Block #1 Final = 0.421 J

KE Block #2 Final = 0.5 * (2.4 kg) * (5.33 m/s)^2
KE Block #2 Final = (1.2 kg) * (28.41 m^2/s^2)
KE Block #2 Final = 34.09 J

Total energy of final system = 0.421 J + 34.09 J = 34.52 J

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