**Ball 1, with a mass of 100 g and traveling at 15 m/s, collides head on with ball 2, which has a mass of 350 g and is initially at rest.**

**What are the final velocities of each ball if the collision is perfectly elastic?**

**Known variables:**

**Mass(m1)** of the first ball = 100 g or .100 kg

**Initial velocity (v1) **of the first ball= 15.0 m/s

**Mass(m2) **of the second ball = 340 g or .340 kg

**Initial velocity(v2) **of the second ball= 0 (initially at rest)

**(V( _{f}x)1 and (V_{f}x)2 **are the velocities of the 2 after the collision

If elastic then the formulas would be the following:

**V(**_{f}**x)1**= 2m_{2}v_{2}/m_{1}+m_{2} + (m_{1}-m_{2}/m_{1}+m_{2})v_{1}

**V(**_{f}**x)2**= 2m_{1}v_{1}/m_{1}+m_{2} + (m_{1}-m_{2}/m_{1}+m_{2})v_{2}

**V(**_{f}**x)1:**

2(.340 x 0)/.100 + .340 + (.100 – .340/ .100 + .340) x (15)

0 + (-.240/.440) x 15

**V(**_{f}**x)1**= -8.3 m/s

**V(**_{f}**x)2**:

2(.100 x 15)/.100 + .340 + (.100 – .340/ .100 + .340) x (0)

2(1.5)/.440 + 0

**V(**_{f}**x)2**= 6.7 m/s

3**) What are the final velocities of each ball if the collision is perfectly inelastic?**

**Inelastic Collision:**

If the collision is inelastic, the combined speed of both balls after the collision can be figured out through this equation.

**v= **m_{1}v_{1}+ m_{2}v_{2}/ m_{1} + m_{2}

(.100 kg x 15 m/s) + 0 / (.100 + .340)

1.5 / .440

=3.3m/s

**V(**_{f}**x)1 & 2 =** 3.3m/s

*The final velocities of each ball would be the same.*

## Leave a Reply