Question

Mass m = 0.1 kg moves to the right with speed v = 0.31 m/s and collides with an equal mass initially at rest. After this inelastic collision the system retains a fraction = 0.69 of its original kinetic energy. What is the speed VR (in m/s) of the mass which was originally at rest before the collision?

Answer 1

VR = 0.26 m/s

Step-by-step explanation:

m = 0.1 kg M = 0.1 kg

v = 0.31 m/s Vi = 0 m/s

the kinetic energy of the system initially is:

Ki = 1/2×m×(v^2) + 1/2×M×(Vi)^2

= 1/2×(0.1)×(0.31)^2

= 4.805×10^-3 J

then, we told that the system after collision only retains a fraction 0.69 of its initial kinetic energy. that is the final kinetic energy of the system is:

Kf = 0.69×4.805×10^-3 J

= 3.31545×10^-3 J

but due equal masses of the bodies, we know that after the collision the only body that would be in motion would be the body that at res and the body that was initially moving will now be at rest.so the kinetic energy is only made by the second body and given by:

Kf = 1/2×M×(VR)^2

3.31545×10^-3 = 1/2×(0.1)×(VR)^2

VR^2 = 0.066309

VR = 0.26 m/s

according to the coservation of linear momentum: