A 3.7 kg block moving with a velocity of +3.5 m/s makes an elastic collision wit

Question

A 3.7 kg block moving with a velocity of +3.5 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 3.7 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. (initially for the 3.7 kg block) (initially for the 2.4 kg block) (finally for the 3.7 kg block) (finally for the 2.4 kg block) Are the two total kinetic energies the same? Yes No

Explanation / Answer

a) Let the velocity after collision be v1 and v2

By the conservation of momentum we have

3.7*3.5 = 3.7v1 + 2.4v2

3.7v1 + 2.4v2 = 12.95 ---------(1)

Velocity of impact = Velocity of recess

v2 - v1 = u1 -u2

v2 - v1 = 3.5 -------(2)

multiply (2) by 3.7 and adding we have

6.1v2 = 25.9

v2 = 4.24 m/s

v1 = v2 - 3.5 = 4.24 - 3.5 = 0.75 m/s

velocity of 3.7 kg block = 0.75 m/s

velocity of 2.4 kg block = 4.24 m/s

b) Initial KE of 3.7 kg block = 0.5*3.7*3.5^2 = 22.66 J

Initial KE of 2.4 kg block = 0

Final KE of block 3.7 kg = 0.5*3.7*0.75^2 = 1.04 J

Final KE of block 2.4 Kg = 0.5*2.4*4.25^2 = 21.53 J

Yes the total kinetic energy is same

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