A single roller-coaster car is moving with speed v0 on the first section of track when it descends a 4.7-m-deep valley, then cli
mbs to the top of a hill that is 5.4 m above the first section of track. Assume any effects of friction or of air resistance are negligible.
(a) What is the minimum speed v0 required if the car is to travel beyond the top of the hill?
(b) Can we affect this speed by changing the depth of the valley to make the coaster pick up more speed at the bottom?
(a) What is the minimum speed v0 required if the car is to travel beyond the top of the hill?
(b) Can we affect this speed by changing the depth of the valley to make the coaster pick up more speed at the bottom?
1 answer:
Explanation:
From the curve given in the attachment,
a) minimum speed v0 required if the car is to travel beyond the top of the hill
climb to the top = 5.4 m
Therefore,
= 10.29 m/s
b) No, we cannot affect this speed by changing the depth of the valley to make the coaster pick up more speed at the bottom as this speed does not depend upon depth of the valley.
You might be interested in
Answer:
24.084 m/s
Explanation:
From the law of conservation of linear momentum
Total momentum before collision equals to the total momentum after collision
Since momentum=mv where m is mass and v is velocity
where
is the mass of the truck,
is velocity of the truck,
is the common velocity of moving and standing truck after collision and
is the mass of the standing truck
Making the subject we obtain
Substituting as 25000 Kg,
as 22.3 m/s,
as 2000 Kg we obtain
Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive
The truck was moving at 24.084 m/s