Question

A metal car starts from rest and slides 12 m down a chute that is inclined at 40° to the horizontal. It then continues to slide along a horizontal length of the chute and eventually comes to rest. If μ = 0,2 between the chute and the car, calculate the horizontal length of the chute. Use conservation of energy to solve this question.

Clearly show the following values in their respective sections

Section A:

E_k=?

E_p=?

E_TR=?

E_TE=?

Energy equation=??

v=??

Section B:

E_k=?

E_p=?

E_TR=?

E_TE=?

Distance=??

Answer 1

given data

$mass of truck = m_{t} = 8000 kg$

$speed of truck =V_{t} = 4 m/s$

$mass of car = m _{c} = 2000 kg$

$speed of car = V _{c} = 16 m / s$

so 

$V_{c} = -16cos 30\degree i + ( -16 sin 30\degree) j$

$$V_{t} = 4i$$

1 

- total momentum before collision 

$p = m_{t} v_{t}+ m _{c} v_{c} = 800 (4 i) + 2000(-16(0.866)i-8j)$

$p = 32000 i - 27712 i - 16 000j$

$p = 4288 i - 16000j$

$|p| = 16564 m kg/s$

2-

as after collision they move together & as momentum is conserve

$|p| = |p_{f}| = 16564 = (m _{t} + m_{c}) v _{f}$

$v_{f} = p/m = 0.04288i - 1.6 j$

$KE _{1} + KE _{2} = m_{t} v_{t}^2/2 + m_{c} v_{c}^2 /2 = E_{i}$

$E_{i} = 4000 (4)^2 + 1000(16)^2$

$E_{i} = 320000 J$

final energy = $(m_{t} + m_{c} /2) v_{f}^2$

$v_{e}^2 =5000(1.66)^2$

$E_{f}=13778 J$

as $E_i$ is not equal to $E_f$ thus not elastic collision