Mechanics ( one easy Questionw hich i cant solve ) A-Levels Sylabbus

[Szeki]

1 a) A particle P of mass m is placed on a rough plane inclined at an angle tan-1 (5/12) to the horizontal.The coefficient of friction between the plane and P is 0.5. Prove that P will remain stationary.

1b ) A light inextensible string is fastened to P, passes up a line of greatest slope, over a frictionless pulley at the top of the plane, and to its other end is attached a particle Q of mass 2m , which hangs freely. Prove that the particles will move and find magnitude of their acceleration.

1c ) When Q has decended a distance h , it hits the floor and rebounds with 0.5 of its speed. Show that Q will hit the floor again before P comes to instantaneous rest.

I can't solve the problem of 1C ) Someone pleae help.
No diagram is included in the question orginally.
Answers for
6b) a= 5g/13 = 3.846 ms-2
6c) T(Q) = 1.387(h)^0.5 Means Root h
T(P)=0.3396(h)^0.5 = 1.074(h/g)^0.5
So t(Q) < t(P)

PLEASE HELP!

 

[barob1n]

The forces acting on Q are gravity and the tension force from being connected to that other mass (and is directed opposite to the gravitational force). The sum of those two forces is equal to the mass of Q (2m) times the acceleration. Solve for the acceleration. Knowing the acceleration and the height that Q falls you can find the speed at which it hits the ground.

The problem then tells you that it rebounds at one half that speed. Find the height that it will reboud to and the time it will take to get there (the mass P should not affect this since it is not pulling on Q). Then find the time it will take to fall from that rebound height (it will be the same time that it took to rebound to that height).

At some point you got to find the time it takes for P to stop moving. To do this you calculate the time it will take for the fricional force plus the gravitational force (remembering it is on an inclined plane) to stop it from a velocity equal the velocity of Q right before it hit the ground for the FIRST time.

Longest answer ever.

 

[kyle8921]

I would first like to commend you for seeking help and clarification on this problem. It is important to reach out when facing difficulties in understanding concepts.

Now, let's break down the problem into smaller parts and see if we can solve it together.

1a) In order for particle P to remain stationary, the forces acting on it must be balanced. These forces include the weight of the particle (mg), the normal force from the plane (N), and the frictional force (F). Since the plane is inclined, we need to resolve these forces along the plane's surface and perpendicular to it.

Along the plane's surface, we have the weight of the particle (mg) acting downward and the frictional force (F) acting upward. We can find the magnitude of the frictional force by multiplying the coefficient of friction (0.5) by the normal force (N) which is equal to mgcos(θ) where θ is the angle of inclination. So, F = 0.5mgcos(θ).

Perpendicular to the plane, we have the normal force (N) acting upward and the weight of the particle (mg) acting downward. Since the plane is inclined at an angle of tan-1(5/12), we can find the value of cos(θ) by using the Pythagorean theorem. So, cos(θ) = 12/13.

Now, let's equate the forces along the plane's surface and perpendicular to it to find the value of N.

Along the plane's surface: mg = F
mg = 0.5mgcos(θ)
N = mgcos(θ)

Perpendicular to the plane: N = mg
mgcos(θ) = mg
cos(θ) = 1

This tells us that the plane must be inclined at an angle of 0 degrees, which means it is a horizontal plane. Therefore, P will remain stationary as there is no inclination for it to slide down.

1b) Now, let's consider the situation with the string and particle Q. Since the pulley is frictionless, the tension in the string is constant throughout. This means that the forces acting on Q are the tension (T) pulling it upward and the weight of the particle (2mg) pulling it downward.

Using Newton's second law, we can write the following equation:

ΣF = ma
T - 2mg = 2