Yatayyat
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I have been doing a few mechanic questions but I have few doubts on some of them and I would like someone to check if I have answered it correctly or not

1) A ship is towed into port by two tugs at a steady speed and course, with each tug pulling on a steel cable. If each cable is under a tension of 9378 N and the angle between them is 44 degrees, Calculate the drag force (in kN) on the ship.

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2) A tennis ball of mass 0.058 kg is projected directly towards a brick wall at 32.4 ms-1. It rebounds with the same speed in the opposite direction.

a) Calculate the magnitude of the impulse on the tennis ball (in Ns) as a result of the collision with the wall.

b) If the collision with the wall lasts 0.37 s, calculate the average magnitude (in N) of the force acting on the ball during the collision.

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3) A car of mass 715 kg is fitted with a trailer of mass 100kg. Assuming a force of 1367 N acts forwards on the car

a) calculate the time taken to accelerate both car and trailer to 22 ms-1 from rest.

b) calculate the tension (in N) in the coupling during acceleration.

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4) A railway truck, mass 1126 kg, is travelling at 16.8 ms-1 towards two identical trucks moving at 3.5 ms-1 in the opposite direction. The three trucks couple together on colliding.

a) Calculate the common speed (in ms-1) of the three trucks after the collision.

b) Calculate the loss of Kinetic Energy (in kJ) as a result of the collision.

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Last edited by Yatayyat; 6 days ago
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ghostwalker
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(Original post by Yatayyat)
...
There's a lot there and I don't have a great deal of time, so will just look at the first one.

Your answer is correct, but there is a far easier method.

Consider the components of the forces parallel and perpendicular to the line you've marked "x".

By symmetry the perpendicular forces cancel out.

And the paralllel forces add to 2\times 9378\times \cos 22 which equals 17390.
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Yatayyat
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(Original post by ghostwalker)
There's a lot there and I don't have a great deal of time, so will just look at the first one.

Your answer is correct, but there is a far easier method.

Consider the components of the forces parallel and perpendicular to the line you've marked "x".

By symmetry the perpendicular forces cancel out.

And the paralllel forces add to 2\times 9378\times \cos 22 which equals 17390.
Yes! Don’t know why I didn’t see that when I was doing this problem, saves a lot more effort from writing more. Thanks
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mqb2766
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(Original post by Yatayyat)
Yes! Don’t know why I didn’t see that when I was doing this problem, saves a lot more effort from writing more. Thanks
Thr other 3 look fine as well. Ive not typed the numbers into a calculator, but they seem ok with a bit of mentall maths.
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Yatayyat
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(Original post by mqb2766)
Thr other 3 look fine as well. Ive not typed the numbers into a calculator, but they seem ok with a bit of mentall maths.
Ok thanks for checking. I redone the numbers on a calculator to make sure everything worked out to be the same still which it is. Appreciate it
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mqb2766
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(Original post by Yatayyat)
Ok thanks for checking. I redone the numbers on a calculator to make sure everything worked out to be the same still which it is. Appreciate it
One thing thats worth doing is to spot obvious cancellations. The carriage momentum one is an obvious example as the masses are all the same so the final velocity is simply the average of the initial velocities. Quicker and cuts down on potential errors.
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Yatayyat
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(Original post by mqb2766)
One thing thats worth doing is to spot obvious cancellations. The carriage momentum one is an obvious example as the masses are all the same so the final velocity is simply the average of the initial velocities. Quicker and cuts down on potential errors.
I'll definitely try to remember to use these quick cancelling outs. And I can see it does work out since for the momentum one as I have 'm(16.8) + m(–3.5) + m(–3.5) = 3m(v)', crossing out the m's on both sides gives '3v = 16.8 – 3.5 – 3.5' so v = 3.3 m/s.
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alekos56
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Why is there a loss in KE. Being a closed system I would have thought that Energy is conserved.
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mqb2766
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(Original post by alekos56)
Why is there a loss in KE. Being a closed system I would have thought that Energy is conserved.
Generally energy isn't conserved for an impact.
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