Question

A 26.5-g object moving to the right at 20.5 cm/s overtakes and collides elastically with a 12.5-g object moving in the same direction at 15.0 cm/s. Find the velocity of each object after the collision. (Take the positive direction to be to the right. Indicate the direction with the sign of your answer.)

Answer 1

v₁ =0.19 m/s and v₂ = 0.18 m/s

Step-by-step explanation:

By conservation of energy and conservation of momentum we can find the velocity of each object after the collision:

Momentum:

Before (b) = After (a)

`m_(1)v_(1b) + m_(2)v_(2b) = m_(1)v_(1a) + m_(2)v_(2a)`

`26.5 kg*0.205 m/s + 12.5 kg*0.150 m/s = 26.5 kg*v_(1a) + 12.5 kg*v_(2a)`
`7.31 kg*m/s = 26.5 kg*v_(1a) + 12.5 kg*v_(2a)` (1)

Energy:

Before (b) = After (a)

`(1)/(2)m_(1)v_(1b)^(2) + (1)/(2)m_(2)v_(2b)^(2) = (1)/(2)m_(1)v_(1a)^(2) + (1)/(2)m_(2)v_(2a)^(2)`

`26.5 kg*(0.205 m/s)^(2) + 12.5 kg*(0.150 m/s)^(2) = 26.5 kg*v_(1a)^(2) + 12.5 kg*v_(2a)^(2)`

`1.40 J = 26.5 kg*v_(1a)^(2) + 12.5 kg*v_(2a)^(2)` (2)

From equation (1) we have:

`v_(2a) = (7.31 kg*m/s - 26.5 kg*v_(1a))/(12.5 kg)` (3)

Now, by entering equation (3) into (2) we have:

`1.40 J = 26.5 kg*v_(1a)^(2) + 12.5 kg*((7.31 kg*m/s - 26.5 kg*v_(1a))/(12.5 kg))^(2)` (4)

By solving equation (4) for
`v_(1a)`, we will have two values for

`v_(1a) = 0.16`

`v_(1a) = 0.21`

We will take the average of both values:

`v_(1a) = 0.19 m/s`

Now, by introducing this value into equation (3) we can find
`v_(2a)`:

`v_(2a) = (7.31 kg*m/s - 26.5 kg*0.19 m/s)/(12.5 kg)`

`v_(2a) = 0.18 m/s`

Therefore, the velocity of object 1 and object 2 after the collision is 0.19 m/s and 0.18 m/s, respectively.

I hope it helps you!