Question

On a frictionless horizontal air table, puck A (with mass 0.250 kg) is moving toward puck B (with mass 0.366 kg), which is initially at rest. After the collision, puckA has velocity 0.116 m/s to the left, and puck B has velocity 0.652 m/s to the right.

Answer 1

momentum equation gives
.250*v=-.250*.116+.366*.652

.250v=-.029+.238
v=.836 m/s to right

kinetic energy of the system before collision=.5*.250*.836^2=.087362
kinetic energy of the system after collision=.5*.250*.116^2 + .5*.366*.652^2
= .001682+ .07779=.07947

0.00788 is the loss in kinetic energy which is obtained by subtracting the kinetic energies before and after collision

Answer 2

From Law of Cons of Momentum:

0.25 * v_ai + 0 = (0.25)*(-.116) + (0.366)*(.652)

Hence v_ai = .209/.25 = .836 m/s

Initial KE: 1/2 * 0.25 * .836^2 + 0 = 0.087

Final KE: 1/2*0.25*(-.116)^2 + 1/2 * 0.366 * (.652)^2 = 0.0795

Difference: 0.0075

Answer 3

Initial momentum of the system
= m(A)*v_ai
Final momentum of the system
= m(A)*v_af + m(B)*v_bf
by momentum conservation we have
Initial momentum of the system = Final momentum of the system
=>.25*v_ai = -.25*0.116 + .366*.652
=> v_ai = 0.2096/.25 = 0.834m/sec

Change in kinetic energy
= .5m(A)*v_ai^2 - .5m(A)v_af^2 - .5m(B)v_af^2
= .5*.25*0.834^2 - .5*.25*.116^2 - .5*.366*.652^2
= 0.0075 J