Given:
v1=8m/s
m1=4.0kg
v2=0
m2=12kg
(a) e=1
∣v1′−v2′∣=∣v1−v2∣=8
Hence
m1v1=m1v1′+m2(8−v1′)
4∗8=4v1′+12(8−v1′)v1′=8m/s
v2′=0
The loss of kinetic energy is zero.
(b) e=0.5
∣v1′−v2′∣=0.5∣v1−v2∣=4
Hence
m1v1=m1v1′+m2(v1′−0.5v1)
4∗8=4v1′+12(v1′−4)v1′=5m/s
v2′=1m/s
The loss of kinetic energy
ΔE=4∗82/2−(4∗52/2+12∗12/2) ΔE=72J
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