Answer :
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)
[tex] m_{1}v_{1b} + m_{2}v_{2b} = m_{1}v_{1a} + m_{2}v_{2a} [/tex]
[tex] 26.5 kg*0.205 m/s + 12.5 kg*0.150 m/s = 26.5 kg*v_{1a} + 12.5 kg*v_{2a} [/tex] [tex] 7.31 kg*m/s = 26.5 kg*v_{1a} + 12.5 kg*v_{2a} [/tex] (1)
Energy:
Before (b) = After (a)
[tex] \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} [/tex]
[tex] 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} [/tex]
[tex] 1.40 J = 26.5 kg*v_{1a}^{2} + 12.5 kg*v_{2a}^{2} [/tex] (2)
From equation (1) we have:
[tex] v_{2a} = \frac{7.31 kg*m/s - 26.5 kg*v_{1a}}{12.5 kg} [/tex] (3)
Now, by entering equation (3) into (2) we have:
[tex] 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} [/tex] (4)
By solving equation (4) for [tex]v_{1a}[/tex], we will have two values for
[tex] v_{1a} = 0.16 [/tex]
[tex] v_{1a} = 0.21 [/tex]
We will take the average of both values:
[tex] v_{1a} = 0.19 m/s [/tex]
Now, by introducing this value into equation (3) we can find [tex]v_{2a}[/tex]:
[tex]v_{2a} = \frac{7.31 kg*m/s - 26.5 kg*0.19 m/s}{12.5 kg}[/tex]
[tex] v_{2a} = 0.18 m/s [/tex]
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!