Math Is Fun Forum

'Weights of 3 kg and 4 kg are connected by a light inextensible string, which passes over a smooth peg. The system is released from rest. Find the acceleration of the system.
After two seconds, the heavier mass strikes the floor and comes to an instantaneous halt. Find the time before the string is again taut.'

I've worked out the acceleration to be 1.4 m/s/s but I'm not sure about the second part. Can somebody help please?

'A small smooth sphere moves on a horizontal table and strikes an identical sphere lying at rest on the table at a distance d from a vertical wall, the impact being along the line of centres and perpendicular to the wall. Prove that the next impact between the spheres will take place at a distance

           2de^2/(1 - e^2)

from the wall, where e is the coefficient of restitution for all the impacts involved.'

Before I tried the problem, I tried the formula on an example:
Let:
Ua be the initial speed of sphere a
Ub be the initial speed of sphere b
Va be the speed of sphere a after the first collision with sphere b
Vb be the speed of sphere b after the first collision
I let Ua = 2 m/s, d = 15 m and e = 1/2 and these are my workings:

by restitution:
1/2 = (Vb - Va)/2
Vb - Va = 1

by conservation of momentum:
2 = Va + Vb

so: Vb = 3/2 m/s and Va = 1/2 m/s

This means that sphere b will take 10 seconds to hit the wall and, during this time, sphere a will have travelled 5 m towards the wall. After the impact with the wall, sphere b will rebound with speed 3/2 x 1/2 = 3/4 m/s. Since there is now 10 m between the spheres and they are heading towards each other: 1/2t + 3/4t = 10 where t is time in seconds. This means they will collide after another 8 seconds and this will be at a distance of 6 m from the wall. Using the formula, however, I get 10 m. What am I doing wrong?

I came across this problem the other day. I understand the vectors behind it but don't really have a definite strategy in solving it. My approach is a bit random. Can someone please advise me of the best way to approach it?

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In the diagram, vector OB = b, vector OC = 4b/3 and vector AP = 2/3 vector AB.
Given that vector AQ = m vector AC and that vector OQ = n vector OP, calculate the values of m and n, and the ratio of the vectors AQ:QC.
I can't see how to use the vectors given only and I can't see any similar triangles. Can somebody get me started?

5 N
     \         |
         \60 deg
              \|________
               |              |
  4 N ----- |________|

A rectangular block B of weight 12 N lies in limiting equilibrium on a horizontal surface. A horizontal force of 4 N and a coplanar force of 5 N inclined at 60 degrees to the vertical act on B.

(i) Find the coefficient of friction between B and the surface.

B is now cut horizontally into two smaller blocks. The upper block has weight 9 N and the lower block has weight of 3 N. The 5 N force now acts on the upper block and the 4 N force now acts on the lower block (see diagram below)

5 N
     \         |
         \60 deg
              \|________
               |_9 N____|
  4 N ----- |_3 N____|

The coefficient of friction between the two blocks is u.

(ii) Given that the upper block is in limiting equilibrium, find u.

(iii) Given instead that u = 0.1, find the accelerations of the two blocks.

my answers:

(i) (4 + 5 sin 60) / (12 - 5 cos 60) = 0.877
(ii) (5 sin 60) / (9 - 5 cos 60) = 0.666
(iii) for the upper block: 5 sin 60 - 0.1(9 - 5 cos 60) = 9a/9.8
so a = 4.01 m/s/s

I don't know how to calculate the acceleration for the lower block as I don't know how the friction between the two blocks affects the tractive force of the lower block. Can somebody help me?

| O | pulley
|    |
|    |
!    !
O  O
A B
Acceleration = 1.4 m s^-2 downwards from particle A

Particles A and B are attached to the ends of a light inextensible string. The string passes over a smooth fixed pulley. The particles are released from rest, with the string taut, and A and B at the same height above a horizontal floor (see terrible diagram). In the subsequent motion, A descends with acceleration 1.4 m s^2 and strikes the floor 0.8 s after being released. It is given that B never reaches the pulley and before A strikes the floor the tension in the string is 5.88 N.

(a) Calculate the mass of A and the mass of B.
(b) The pulley has mass 0.5 kg, and is held in a fixed position by a light vertical chain. Calculate the tension in the chain immediately before and after A strikes the floor.

(a) for A: mg - 5.88 = 1.4ma so m = 0.7 kg
     for B: 5.88 - mg = 1.4ma so m = 0.525 kg

I need help with (b). I would've thought we could just take the weight of the particles and the pulley, i.e. (0.5 + 0.7 + 0.525) x 9.8 = 16.905 N but this doesn't equate to the sum of the weight of the pulley and the tensions in the string, i.e. 0.5 x 9.8 + 2 x 5.88 = 16.66 N
Why is there a difference?