A particle of mass 0.5 kg is held at rest at a point P, which is at the bottom of an inclined plane. The particleis given an impulse of 1.8 N s directed up a line of greatest slope of the plane. (i) Find the speed at which the particle starts to move.
The particle subsequently moves up the plane to a point Q, which is 0.3 m above the level of P. (ii) Given that the plane is smooth, find the speed of the particle at Q.
The initial velocity is 3.6ms. I found the answer to question 2 by subtracting the mgh from the initial K, as such: 1/2mv^2 = 1/2mu^2 - mgh; and solving for v. V= 2.66 ms.
I do not understand how I got this answer, because the particle is travelling up the slope so should be losing KE and gaining PE, so it makes more sense to me to subtract the KE instead of the PE. I would really appreciate some help. Thanks in advance.
Mechanics Help Watch
- Thread Starter
- 14-05-2016 22:00
- 14-05-2016 23:45
Hi I'm not doing mechanics 2 but thought I would have a look. The ball is moving up the slope with a given kinetic energy, which at point Q the initial kinetic energy supplied to the ball will be reduced and the gravitational potential will be increased (conservation of energy).
Therefore kinetic energy before, which I worked out to be 1/2*0.5*3.6^2 = 3.24J is equal to the sum of the final gravitational potential energy (mgh) and the final kinetic energy.
3.24 = 1/2*m*v(final)^2 + mgh
therefore v(final) = 2.66m/s