Two pucks are directed towards each other on an ice rink (along the x-axis.) Puc

Question

Two pucks are directed towards each other on an ice rink (along the x-axis.) Puck has a mass of 0.16 kg and is moving with a speed of 4 m/s in the positive x-direction. Puck B has a mass of 0.17 kg and is moving with a speed of 7 m/s in the negative x-direction. They collide nearly elastically (no energy is dissipated to the heating of the pucks or the ice and the pucks don't deform too much). Puck A slides away with a speed of 7.33 m/s in the negative x-direction. Determine the velocity of Puck B after the collision. To solve part (a), what did you include in your system? What did you assume that the total force due to the surroundings on each puck was during the collision? The collision is very short; it occurs over 0.036s. Determine the average force that Puck A exerts on Puck B. Determine the average force that Puck B exerts on Puck A. Determine the average acceleration of Puck B during the collision. Determine the average acceleration of Puck A during the collision.

Explanation / Answer

a)

Initial momentum of the system:

Pi = 0.16*4 - 0.17*7 = -0.55 kg.m/s

Final momentum of the system:

Pf = 0.16*(-7.33) + 0.17*v

By conservation of momentum,

Pi = Pf

So,  -0.55 = 0.16*(-7.33) + 0.17*v

So, v = 3.66 m/s <-------answer

b)

We include only Puck A and Puck B into system .

c)

Total force due to the surrounding = 0 N <------ as there is no friction

So, answer : ( 0,0,0 ) N

d)

change of momentum of Puck B:

Pb = mB*(vfB - viB)

= 0.17*((3.66) - (-7))

= 1.81 kg.m/s

So, average force = Pb/t

= 1.81/0.036

= 50.3 N

e)

Similarly average force on A:

Fa = 0.16*(-7.33 - 4 )/0.036 = -50.3 N

f)

average acceleration of Puck B:

aB = Fb/m = 50.3/0.17 = 295.9 m/s2

g)

aA = Fa/m = -50.3/0.16 = -314.4 m/s2

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