Mass m = 0.1 kg moves to the right with speed v = 0.39 m/s and collides with an equal mass initially at rest. After this inelastic collision the system retains a fraction = 0.86 of its original kinetic energy. If the masses remain in contact for 0.01 secs while colliding, what is the average force in N between the masses during the collision? Hints: All motion is in 1D. Ignore friction between the masses and the horizontal surface. You will probably need to use the quadratic formula to solve the resulting equations. VR must be greater than VL since the masses can't pass through each other!
Mass m1 = 0.1 kg & Mass m2 = 0.1 kg
intial velocity of mass m1 , u1 = 0.39 m/s
Intial kinetic energy of mass m1 is K1 = 1/2 (m1u12 )= 1/2 x 0.1 x (0.39)2 = 0.007605 Joule
Remaining kinetic energy of mass m1 after collision = 0.86 x intial kinectic energy of mass m1 = 0.86 x 0.007605 = 0.0065403 Joule
then final kinetic energy of mass m1 after collision ,K2 = 0.0065403 Joule
Let final velocity of mass m1 after collision = u2
then using K2 = 1/2 (m1u22 )
0.0065403 = 1/2 x 0.1 x u22
u2 = 0.36167112132
Then change in momentum of mass m1 is dp = m1(u1-u2)
dp = 0.00283288786
contact time dt = 0.01 sec
hence force between masses F = dp/dt = 0.283288786 Newton
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