A 2.8 kg block moving with a velocity of +4.6 m/s makes an elastic collision wit

Question

A 2.8 kg block moving with a velocity of +4.6 m/s makes an elastic collision with a stationary block of mass 2.0 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 2.8 kg block)
__________m/s (for the 2.0 kg block)

(b) Check your answer by calculating the initial and final kinetic energies of each block.
________J (initially for the 2.8 kg block)
__________J (initially for the 2.0 kg block)

________J (finally for the 2.8 kg block)
_________J (finally for the 2.0 kg block)

Are the two total kinetic energies the same? Yes or no?

Explanation / Answer

A. In eleastic collision

u1 - u2 = v2 - v1

u2 = 0

momentum conservation

m1*u1+ m2*u2 = m1v1 + m2v2

solvin these two equations

v1 = (m1 - m2)*u1/(m1+m2) = (2.8 - 2)*4.6/(2.8+2) = 0.8*4.6/4.8 = 0.77 m/sec

v2 = 2*m1*v1/(m1+m2) = 2*2.8*4.6/4.8 = 5.37 m/sec

B. KEi1 = 0.5*m1*u1^2 = 0.5*2.8*4.6^2 = 29.62 J

KEi2 = 0 as u2 = 0

C. KEf1 = 0.5*2.8*0.77^2 = 0.83 J

KEf2 = 0.5*2*5.36^2 = 28.73 J

Yes the two total kinetic energy is same.

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