Question #299244

A body of 4kg mass travelling at 8m/s² collides with a stationary body of mass 12kg, both being constrainted in their motion by the same frictionless guides. Find their velocities and loss of kinetic energy, if coefficient of restitution is 1 and (b) if it is 0.5

Expert's answer

Given:

v1=8 m/sv_1=8\:\rm m/s

m1=4.0 kgm_1=4.0\:\rm kg

v2=0v_2=0

m2=12 kgm_2=12\:\rm kg


(a) e=1e=1

∣v1′−v2′∣=∣v1−v2∣=8|v_1'-v_2'|=|v_1-v_2|=8

Hence


m1v1=m1v1′+m2(8−v1′)m_1v_1=m_1v_1'+m_2(8-v_1')

4∗8=4v1′+12(8−v1′)4*8=4v_1'+12(8-v_1')

v1′=8 m/sv_1'=8\:\rm m/s

v2′=0v_2'=0

The loss of kinetic energy is zero.

(b) e=0.5e=0.5

∣v1′−v2′∣=0.5∣v1−v2∣=4|v_1'-v_2'|=0.5|v_1-v_2|=4

Hence


m1v1=m1v1′+m2(v1′−0.5v1)m_1v_1=m_1v_1'+m_2(v_1'-0.5v_1)

4∗8=4v1′+12(v1′−4)4*8=4v_1'+12(v_1'-4)

v1′=5 m/sv_1'=5\:\rm m/s

v2′=1 m/sv_2'=1\:\rm m/s

The loss of kinetic energy

ΔE=4∗82/2−(4∗52/2+12∗12/2)\Delta E=4*8^2/2-(4*5^2/2+12*1^2/2)

ΔE=72 J\Delta E=72\:\rm J


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