PLEASE HELP ME SOLVE THIS QUESTION.

Question

P is a point 10m from the bottom of a slope, A ball is rolled up the slope from P with as initial velocity of 8 m.s-1 , it undergoes a constant acceleration of -4 m.s-2

 

a-

 i- How long does it take to reach the bottom of the slope ?

 

ii - How far up the slope does it travel ?

 

 

B- One second after the first ball is rolled up from the bottom of the slope it has an initial velocity of 14 m.s-1 . It also has an acceleration of -4 m.s-2 

Find when and where the two balls meet ?

Answer 1

Hi Nona!

 

I take it that the bottom of the slope leads onto a horizontal surface, so that the ball rolled from P rolls horizontally and then onto the slope.  If that is the case, then its horizontal travel from P to the bottom of the slope is (presumably) at constant velocity.  So you would just use the non-acceleration equation of motion to work with it:

 

d = vt

 

where

 

d = displacement

v = velocity

t = time

 

You know d and v on the horizontal surface, and so can solve for t.

 

For the second question, you can treat the direction along the slope as one dimension, and use one-dimensional kinematics along the slope.  For this case, of knowing how far up the slope the ball rolls, we are trying to find displacement ("how far").  So:

 

to find:  d

know:  a (=-4 m/s²), v_o (= 8 m/s, the velocity it has when starting up the slope), v (=0, since the marble must come to a stop at the moment it is as far up the slope as it can go)

 

So, we can use the kinematic equation:

 

v² = v_o²+2ad

 

And solve for d.

 

If the second ball is rolled 1 s after the first balled is rolled from Point P, and the second ball starts at the bottom of the slope, I believe the second ball will start up the slope BEFORE the first ball arrives at the bottom (which should take more than one second rolling from 10 m away at 8 m/s).  In fact, if my calculation is correct, the second ball should start up the slope 0.25 s before the first ball arrives at the bottom from Point P.

 

So, the displacement of the 2nd ball up the slope as a function of time (and other quantities) is given by:

 

d₂ = v_o2t₂+(1/2)at₂² = 14t₂+(1/2)(-4)t₂² = 14t₂ - 2t₂²

 

And we can write a similar expression for the first ball from Point P:

 

d₁ = v_o1t₁+(1/2)at₁² = 10t₁+(1/2)(-4)t₁² = 10t₁-2t₁²

 

In addition, if we start our timer when Ball 1 reaches the bottom of the slope from Point P, Ball 2 will already have been on the slope for 0.25 s.  So it is where it would be 0.25 AFTER t = 0.  This is the same for any point later, when comparing the two.  Ball 2 is always where it would be 0.25 s after the time indicated by t₁ .  Mathematically, this becomes t₂ = t₁+0.25.  We want to put this in for t₂ if referencing our time measurements to t₁

 

So, we need equal displacements, for the two balls to meet, meaning:

 

d₁ = d₂

 

10t₁-2t₁² = 14(t₁+0.25)-2(t₁+0.25)²

 

Since t₁ is the only unknown here, you should be able to solve this equation for it.  That will be the time from when the first ball enters the plane to when they collide.  Putting this back into the expression for d₁ will give the displacement of the plane at which this occurs.

 

Hope this helps!  Let me know if you have any questions, or any other problems with this.  I will also re-check it later for accuracy.