[smith50]

Hi, part b) is where i'm stuck at could anyone explain this i know im partly right but am getting confused in the final steps.
Thanks in advance

[raheem94]

(Original post by smith50)

Hi, part b) is where i'm stuck at could anyone explain this i know im partly right but am getting confused in the final steps.
Thanks in advance

Use the formula e=v/u to find an expression for v. The u here is the velocity of Q after the first collision with P.

Now find an expression for the speed of Q after the second collision with P by using newton's law of restitution, we know that the coefficient between P and Q is 1/3.

Now use conservation of momentum.

Do i make any sense?

Can you post your working, so that i can guide you from the step where you are stuck.

[smith50]

(Original post by raheem94)
Use the formula e=v/u to find an expression for v. The u here is the velocity of Q after the first collision with P.

Now find an expression for the speed of Q after the second collision with P by using newton's law of restitution, we know that the coefficient between P and Q is 1/3.

Now use conservation of momentum.

Do i make any sense?

Can you post your working, so that i can guide you from the step where you are stuck.

Thanks i get it now
Could you help me on this question i have shown my working .Thanks in advance

[raheem94]

(Original post by smith50)
Thanks i get it now
Could you help me on this question i have shown my working .Thanks in advance

You have written the speed of A after the first collision wrong in your working, it should be:

For the last part,
First use conservation of momentum between B and C.
Then use restitution law between B and C.
Now solve the simultaneous equations to eliminate and get an equation which links and . Now make the subject of this equation, you know the expression for , substitute this in the equation. This will give you an expression for in terms of 'u' and 'e'.

Now to show that there will be another collision solve, .
The 'u's' will cancel out and you will get an expression in terms of 'e'. Now see if the range given by it is correct or not, if its correct than there will be another collision.

[smith50]

(Original post by raheem94)
You have written the speed of A after the first collision wrong in your working, it should be:

For the last part,
First use conservation of momentum between B and C.
Then use restitution law between B and C.
Now solve the simultaneous equations to eliminate and get an equation which links and . Now make the subject of this equation, you know the expression for , substitute this in the equation. This will give you an expression for in terms of 'u' and 'e'.

Now to show that there will be another collision solve, .
The 'u's' will cancel out and you will get an expression in terms of 'e'. Now see if the range given by it is correct or not, if its correct than there will be another collision.

Majorly stuck on this quetion could anyone explain to me.

[ghostwalker]

(Original post by smith50)
Majorly stuck on this quetion could anyone explain to me.

What's the problem? You presumably know how to break down a shape, or in this case consider it as one shape, with another subtracted from it.

Also note the line of symmetry.

[smith50]

(Original post by ghostwalker)
What's the problem? You presumably know how to break down a shape, or in this case consider it as one shape, with another subtracted from it.

Also note the line of symmetry.

For the two triangles i dont know how to calculate the centre of mass coordinates and line of symmetry how so x bar and y bar will be the same????

[ghostwalker]

(Original post by smith50)
For the two triangles i dont know how to calculate the centre of mass coordinates

Centre of a mass of a triangle is one third the distance up any altitude. In the case of a right angled triangle, the important ones are: This will be 1/3 the distance along the two sides at the right angle, working from the right angle.

and line of symmetry how so x bar and y bar will be the same????

Yes; if you're treating B as the origin, and sides BA, BE as the axes, since BE is the line of symmetry.

[raheem94]

(Original post by smith50)
Majorly stuck on this quetion could anyone explain to me.

Consider the 2 triangles separately, find the area of both of them.

Find the centre of mass of each triangle, by using the coordinates marked in the diagram.

At the end subtract them,