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

Answer:

v₁ =0.19 m/s and v₂ = 0.18 m/s

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)                        

$\frac{1}{2}m_{1}v_{1b}^{2} + \frac{1}{2}m_{2}v_{2b}^{2} = \frac{1}{2}m_{1}v_{1a}^{2} + \frac{1}{2}m_{2}v_{2a}^{2}$          

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

$1.40 J = 26.5 kg*v_{1a}^{2} + 12.5 kg*v_{2a}^{2}$   (2)    

From equation (1) we have:

$v_{2a} = \frac{7.31 kg*m/s - 26.5 kg*v_{1a}}{12.5 kg}$   (3)

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

$1.40 J = 26.5 kg*v_{1a}^{2} + 12.5 kg*(\frac{7.31 kg*m/s - 26.5 kg*v_{1a}}{12.5 kg})^{2}$    (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 $v_{2a}$:

$v_{2a} = \frac{7.31 kg*m/s - 26.5 kg*0.19 m/s}{12.5 kg}$

$v_{2a} = 0.18 m/s$

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!