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# Need help with these questions!! watch

1. Need help with these mock exam questions.

1. A box is pulled forward along a frictionless horizontal surface using a rope angled at 20° above the horizontal. The force pulling the block is 250N along the handles. Draw a suitable diagram showing the force in the handles, and its horizontal and vertical components, and use trigonometry to calculate the force parallel to the surface. What is the force in a direction perpendicular to the surface?

2. A 90 kg box is placed on a frictionless track inclined at 25° to the horizontal. Draw a diagram of this identifying the relevant forces, e.g force due to gravity, forces normal and tangential to the track. What is the magnitude of the force on the box normal to the track? Calculate the tangential force and hence the acceleration of the box down the track.

3. A van of mass 1750kg is travelling along a road at 23m/s along a road when it runs into the back of a car of mass 800kg that is travelling in the same direction at 15m/s. After the collision the 1750kg van travels at a speed of 19m/s in the same direction. Use the principle of the conservation of momentum to calculate the speed of the 800kg car after the collision

What is the combined kinetic energy of the boats before and after collision? Is Kinetic Energy conserved?

4. A rollercoaster carriage weighing 225kg travels down a frictionless roller starting at a height of 18m above the ground. The rollercoaster track goes down to 1m above the ground as it curves back up to a peak at 10m before falling again

Using the conservation of mechanical energy, determine the kinetic energy of the carriage at the lowest point after the start. How fast is the sack travelling when it travels through the lowest point?

How fast will it be travelling as it passes over the next peak?
2. (Original post by Muss963)
Need help with these mock exam questions.

1. A box is pulled forward along a frictionless horizontal surface using a rope angled at 20° above the horizontal. The force pulling the block is 250N along the handles. Draw a suitable diagram showing the force in the handles, and its horizontal and vertical components, and use trigonometry to calculate the force parallel to the surface. What is the force in a direction perpendicular to the surface?

2. A 90 kg box is placed on a frictionless track inclined at 25° to the horizontal. Draw a diagram of this identifying the relevant forces, e.g force due to gravity, forces normal and tangential to the track. What is the magnitude of the force on the box normal to the track? Calculate the tangential force and hence the acceleration of the box down the track.

3. A van of mass 1750kg is travelling along a road at 23m/s along a road when it runs into the back of a car of mass 800kg that is travelling in the same direction at 15m/s. After the collision the 1750kg van travels at a speed of 19m/s in the same direction. Use the principle of the conservation of momentum to calculate the speed of the 800kg car after the collision

What is the combined kinetic energy of the boats before and after collision? Is Kinetic Energy conserved?

4. A rollercoaster carriage weighing 225kg travels down a frictionless roller starting at a height of 18m above the ground. The rollercoaster track goes down to 1m above the ground as it curves back up to a peak at 10m before falling again

Using the conservation of mechanical energy, determine the kinetic energy of the carriage at the lowest point after the start. How fast is the sack travelling when it travels through the lowest point?

How fast will it be travelling as it passes over the next peak?
What have you tried so far?
3. Try and understand principles and have a shot - make an effort to suppress the encouragement of working like a robot and just entering values into equations (if your school is as bad as some!).

In diagram below (Q2) the force of gravity would be downward 90 X 9.8 N.

Chk out your maths to remind yourself the defintions of sine and cosine. Work out which of those you need to use, respectively, to work out forces (vectors) in the plane parallel to the slope (tangential) and normal to the slope (at 90 degrees).

Q3 Total momentum in direction of travel b4 accident = (1750 X 23) + (800 X15) kg m s^-1

You know the momentum of the van after accident - look up conservation of momentum in your book. Knowing that total momentum does not change, and knowing the mass of the car, work out the speed of the car after the accident.

The other Qs follow similar principles, so u should find them possible to solve thereafter.

GO ON YOUNG MAN/LADY - you can do it!

Mukesh (Science Tutor)
4. (Original post by macpatelgh)
Try and understand principles and have a shot - make an effort to suppress the encouragement of working like a robot and just entering values into equations (if your school is as bad as some!).
...
Sorry, that is just not called for - if you are a tutor then part of that role is to support not criticise ...

5. Need help understanding question 3 and how to solve it !
6. (Original post by Muss963)
Need help understanding question 3 and how to solve it !
at any given instant the total momentum = momentum of car + momentum of van

by conservation of momentum the total momentum the instant before the collision is equal to the total momentum the instant after the collision.

(kinetic energy does not need to be conserved in collisions, don't assume it is unless you are told that they are elastic... but momentum always is conserved)
7. (Original post by Joinedup)
at any given instant the total momentum = momentum of car + momentum of van

by conservation of momentum the total momentum the instant before the collision is equal to the total momentum the instant after the collision.

(kinetic energy does not need to be conserved in collisions, don't assume it is unless you are told that they are elastic... but momentum always is conserved)
So what would be the answer to that question and how would you work it out on paper?
8. before collision

mvan1=1750 kg
vvan1=23 m/s

mcar1=800 kg
vcar1=15 m/s

after collision

mvan2=1750 kg
vvan2=19 m/s

mcar2=800 kg
vcar2... unknown
---
from conservation of momentum
(mvan1 vvan1) + (mcar1 vcar1) = (mvan2 vvan2) + (mcar2 vcar2)

collect the known values on one side of the equals sign
(mvan1 vvan1) + (mcar1 vcar1) - (mvan2 vvan2)=(mcar2 vcar2)

((mvan1 vvan1) + (mcar1 vcar1) - (mvan2 vvan2)) / mcar2 = vcar2

=(40250 + 12000 - 33250)/800

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