Question

A 5.00 g object moving to the right at 20.0 cm/s makes an elastic head-on collision with a 10.0 g object that is initially at rest. (a) Find the velocity of each object after the collision. -6.67 Correct: Your answer is correct. cm/s (5.00 g object) 13.33 Correct: Your answer is correct. cm/s (10.0 g object) (b) Find the fraction of the initial kinetic energy transferred to the 10.0 g object. 88.44 Correct: Your answer is correct. %

Answer 1

a)
`v_(1f)=-6.67\: cm/s`

`v_(2f)=13.33\: cm/s`

b)
`n=88.84\: \%`

Step-by-step explanation:

a) Applying the conservation of momentum, we have:

`p_(i)=p_(f)`

p(i) is the initial momentum. In our case is due to the 5 g object.

p(f) is the final momentum. Here, both objects contribute.

`m_(1i)v_(1i)=m_(1f)v_(1f)+m_(2f)v_(2f)`

Where:

• m(1) is 5 g
• m(1) is 10 g
• v(1i) is the initial velocity 20 cm/s or 0.2 m/s

To find both final velocities we will need another equation, let's use the conservation of kinetic energy.

`m_(1i)v_(1i)^(2)=m_(1f)v_(1f)^(2)+m_(2f)v_(2f)^(2)`

So we have a system of equations:

`5*0.2=5v_(1f)+10v_(2f)` (1)

`5*0.2^(2)=5v_(1f)^(2)+10v_(2f)^(2)` (2)

Solving this system we get:

`v_(1f)=-6.67\: cm/s`

`v_(2f)=13.33\: cm/s`

b) The fraction of the initial kinetic energy transferred is:

`n=(m_(2)v_(2f)^(2))/(m_(1)v_(1i)^(2))`

`n=(10*13.33^(2))/(5*20^(2))`

`n=88.84\: \%`

I hope it helps you!