Answer:
a. 2 kg*m/s
b. [tex]p_{T_{f}} = 0.5v_{f} = 2 kg*m/s[/tex]
c. 4 m/s
Explanation:
a. The momentum of the system ([tex]p_{Ti}[/tex]) before the blob of clay strikes the cart is:
[tex] p_{Ti} = p_{b} + p_{c} [/tex]
Where:
[tex]p_{b}[/tex] is the momentum of the blob clay
[tex]p_{c}[/tex] is the momentum of the car
[tex] p_{Ti} = m_{b}v_{b} + m_{c}v_{c} [/tex]
Since the car is initially at rest, [tex]v_{c}[/tex] = 0
[tex] p_{Ti} = 200 g*\frac{1 kg}{1000 g}*10 m/s + 0 = 2 kg*m/s [/tex]
b. The momentum of the system after they come together:
[tex]p_{T_{f}} = m_{b}v_{b} + m_{c}v_{c}[/tex]
Since they come together, [tex]v_{b}[/tex] =
[tex]p_{T_{f}} = v_{f}(m_{b} + m_{c}) = v_{f}(0.2 kg + 0.3 kg) = 0.5v_{f}[/tex] (1)
Because we do not have the final speed we can not calculate the final momentum.
c. We can find the speed of the clay and car by conservation of the momentum:
[tex] p_{i} = p_{f} [/tex]
The initial momentum of the system was founded in part "a" (p = 2 kg*m/s), so we have:
[tex] 2 kg*m/s = m_{b}v_{b_{f}} + m_{c}v_{c_{f}} [/tex]
Again, when they come together, the final speed is the same:
[tex] 2 kg*m/s = v_{f}(m_{b} + m_{c}) [/tex]
[tex] v_{f} = \frac{2 kg*m/s}{0.2 kg + 0.3 kg} = 4 m/s [/tex]
Now, since we found the final speed we can calculate the momentum of the system after they come together (equation 1):
[tex] p_{T} = 0.5v_{f} = 0.5 kg*4m/s = 2 kg*m/s [/tex]
I hope it helps you!
