All Mechanics DOUBTS HERE!!!

[elemis]

mmm, alright, that makes sense too!

My question was, when calculating the acceleration for A, it was taken as 5/7 and not 1.4 like we had found in part a) of the question, why is that?

In part a tension was an added factor, when the tension disappears the acceleration changes.

M1 is about common sense and spatial awareness.

[elemis]

Okay, another question:

I don't understand how part c) is solved, the working is given below!

Check the attachment, it has the question and the answer!

Note:
Acceleration of the system = 0.613
Tension in the string = 27.6

Consider the two tensions as forces (never mind the direction they act in) and find the resultant force.

[astarmathsandphysics]

My answer

[astarmathsandphysics]

The two tensions are the same because the particle in free to move.

[**RoRo**]

Consider the two tensions as forces (never mind the direction they act in) and find the resultant force.

Alright, thank you very much!

[Zishi]

The two tensions are the same because the particle in free to move.

Alright, thank you! And what about the third part which I attached in my previous post?

EDIT: By the way I've attached my diagram showing what I did, and how I resolved forces, etc. Please tell me why I can't find R, and if is any direction wrong?!

[Sue T]

i have a question in 2007 june

7. Two small spheres P and Q of equal radius have masses m and 5m respectively. They lie
on a smooth horizontal table. Sphere P is moving with speed u when it collides directly
with sphere Q which is at rest. The coefficient of restitution between the spheres is e,
where e > 1/5

(a) (i) Show that the speed of P immediately after the collision is u/6(5e-1)

(ii) Find an expression for the speed of Q immediately after the collision, giving your
answer in the form ?u, where ? is in terms of e.
the marking scheme dint say nything bout this one - wat is it??!

Three small spheres A, B and C of equal radius lie at rest in a straight line on a smooth
horizontal table, with B between A and C. The spheres A and C each have mass 5m, and the
mass of B is m. Sphere B is projected towards C with speed u. The coefficient of restitution
between each pair of spheres is .

(b) Show that, after B and C have collided, there is a collision between B and A.

th marking scheme said:
After B hits C, velocity of B = “v” = 1/6 (1 – 5*4/5 )u = – ½u
velocity < 0 ? change of direction ? B hits A
but were did all the negative come from - wasn't it  u/6(5e-1)? why'd they reverse the signs out of no where??! is it jus 4 the sake of proving tht there'll be a collision?!

[**RoRo**]

How do you solve part c) ?

Check the attachment below for the question.

[elemis]

How do you solve part c) ?

Check the attachment below for the question.

2 minutes.

[**RoRo**]

(a) -2g-0.4(2g*0.8 ) = 2*a

Therefore a = -12.936 ms^-2

Using V^2-U^2=2as we find distance = 0.154607 m ?

Correct ?

Nope, the correct answer is 1.1ms^-1 (2 s.f.)

& the working on the CD is attached, but I don't understand how did the get that 2a = 14g/25?

In part b) we had to show that the box will slide down the plane, so I took the net force down the plane and that was the answer [14g/25], how did that change to acceleration all of a sudden?

[elemis]

Nope, the correct answer is 1.1ms^-1 (2 s.f.)

& the working on the CD is attached, but I don't understand how did the get that 2a = 14g/25?

In part b) we had to show that the box will slide down the plane, so I took the net force down the plane and that was the answer [14g/25], how did that change to acceleration all of a sudden?

I made a mistake using the wrong initial speed.

When the diagram slides back down the plane it will accelerate obviously.

Using the below diagram :

We can see the friction acts in the opposite direction to motion. Resolving the forces :

Hence, 2g*sin alpha -0.4*2g cos alpha = 2*a

a = 2.744 ms^-2

Since the velocity at the top of the slope will be zero :

v^2 = 0^2 + 2*a*0.22

V = 1.10 ms^-1

[elemis]

For part (b) we can imagine the particle momentarily at rest.

Considering the frictional force only : 0.4*R = 6.272 N

Comparing this to the weight : 2g*0.6 = 11.76 N

We can see that friction cannot prevent the weight from pulling the object down.

[**RoRo**]

I made a mistake using the wrong initial speed.

When the diagram slides back down the plane it will accelerate obviously.

Using the below diagram :

We can see the friction acts in the opposite direction to motion. Resolving the forces :

Hence, 2g*sin alpha -0.4*2g cos alpha = 2*a

a = 2.744 ms^-2

Since the velocity at the top of the slope will be zero :

v^2 = 0^2 + 2*a*0.22

V = 1.10 ms^-1

How is s=0.22?
And the acceleration will not be taken as a negative value because when we resolved the forces and found the acceleration, the resultant was in the downwards direction?

[**RoRo**]

Can anyone please solve part b for this question with the steps for the working?

The answer for a) 11.4N and for b) is 13.9N

I found that I can get the answer by doing the following:

R= 11.4Cos35 + 8Cos55 = 13.9

But I don't understand why is it an addition sign?

The horizontal 8N force, when resolved, should have 2 values, one that acts up the plane and the other one is perpendicular to it, why is it pointing downwards? [to get the answer 13.9]

I'm hoping my question is clear, if not, please let me know and I'll clarify further!

[astarmathsandphysics]

look at the arrows on the 8N force on the diagram below. The force is pushing the block into the plane