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A 0.450-kg ice puck, moving east with a speed of 5.16 m/s , has a head-on...

A 0.450-kg ice puck, moving east with a speed of 5.16 m/s , has a head-on collision with a 0.950-kg puck initially at rest. Assume that the collision is perfectly elastic.

What is the speed of the 0.450-kg puck after the collision?

Express your answer to three significant figures and include the appropriate units.

What is the speed of the 0.950-kg puck after the collision?

Express your answer to three significant figures and include the appropriate units.

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Answer #1

Given, the first ice puck has mass (m) = 0.450 kg

the first ice puck has initial velocity (u1) = 5.16 m/s

the first ice puck has mass (M) = 0.950 kg

the first ice puck has initial velocity (u2) = 0 m/s

Let the final velocity of mass mafter collision is v1 and final velocity of mass M after collision is v2 .

Now from conservation of momentum ,

  m u_{1} + M u_{2} = m v_{1}+ M v_{2}

or, 0.450 × 5.16 + 0.950 × 0 0.450u1 + 0.9500g

or,   2.322 299 0.450u1 + 0.950u2

or, 2.322 0.950v2 0.450 ..........................(1)

Here the collision is perfectly elastic. So, in this case coefficient of restitution (e) will be 1.

The coefficient of restitution is defined as   

02 1 u12

for this case

e = 1

or,   uu2

or, 2112

or, 2 -v1- 5.16 -(0

or, v_{2}=5.16 + v_{1}

putting this value in equation (1), we get

  2.322 0.950(5.16 + v) 0.450

or,   2.58 - 0.950vi 0.450

or,   0.450u1 =ー2.58-0.950u1

or, 1.4v1- -2.58

or,   1--1.84 m/s

So,

2-5.16- 1.84-3.32 m/s

Hence, the speed of the 0.450 kg puck after collision is 1.84 m/s in the opposite direction of initial motion (i.e in west direction) and the speed of 0.950 kg puck after collision is 3.32 m/s in the east direction .

For any doubt please comment and please give an up vote. Thank you.

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