Question

In: Physics

A 0.415-kg hockey puck, moving east with a speed of 4.75 m/s, has a head-on collision...

A 0.415-kg hockey puck, moving east with a speed of 4.75 m/s, has a head-on collision with a 0.830-kg puck initially at rest.

Part A

Assuming a perfectly elastic collision, what will be the speed (magnitude of the velocity) of each object after the collision?

Part B

What will be the direction of the lighter object after the collision.

[ ] West

[ ] East

Part C

What will be the direction of the heavier object after the collision.

[ ] West

[ ] East

Solutions

Expert Solution

mass of puck 1 ,m1 =0.415 kg

initial velocity of puck 1 ,u1 =4.75 m/s

mass of puck 2,m2 =0.830 kg

initial velocity of puck 2 ,u2 =0 m/s

----------

Given collision is perfectly elastic

so Momentum is conserved

Momentum before collision =Momentum after collision

-----------------(1)

-------------------------------------

Kinetic Energy is conserved

Kinetic Energy before collision =Kinetic Energy after collision

------------------(2)

------------------------------

Put (1) in (2)

(Multiply by 0.83 kg)

Solve the quadratic equation

we get 2 values ,: 4.8 m/s and -1.58 m/s

Take the negative value sine the positive value is larger than initial velocity of mass 1

ANSWER:

From (1)

ANSWER:

====================

Part B) What will be the direction of the lighter object after the collision.

Negative value so ANSWER: WEST

Part C) What will be the direction of the heavier object after the collision.

Positive value so ANSWER: EAST

==========================


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