# Block off inclined plane (kinematics + projectile motion?)

1. Feb 21, 2016

### muhammed_oli

1. The problem statement, all variables and given/known data
A block of mass m = 2.00 kg is released from rest at h = 0.300 m above the surface of a table, at the top of a θ = 50.0° incline as shown in the figure below. The frictionless incline is fixed on a table of height H = 2.00 m.
a. Find a on the incline
a=7.51 :)

b. Find V(f) on incline
V(f)=2.42 :)

c. How far from the table will the block hit the floor?
:(

d. What time interval elapses between when the block is released and when it hits the floor?
:(

2. Relevant equations
incline kinematics
y=V(o)t-g/2(t^2)

projectile stuff
t=(2V(o)sin∅)/g
range(x)=(V(o)^2(sin2∅))/g

3. The attempt at a solution
It's free fall kinematics for the plane then projectile for when it falls? I was distracted when he was showing us this

c. range(x)=(2.42^2)sin(100)/g

d. -0.300=0-4.9t^2
t(p)=0.247
t(f)=2(2.42)sin(50)/g
t(f)=0.378
t(tot)=0.247+0.378=0.625 sec

I already missed c but is that how i get there?
Part D wrong.

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Last edited: Feb 21, 2016
2. Feb 21, 2016

### haruspex

The range equation you quote for c refers to a certain set-up. What is that set-up? Note that the equation has no input distances, but clearly the height of the table will be important. (It's not much use memorising equations if you don't remember the circumstances in which they apply. I have never bothered memorising many equations, preferring to remember how to use the basic ones.)
So, for c, what are the horizontal and vertical components of velocity when it leaves the ramp?

Your first equation for part d is as though the block falls vertically 0.3m. You already calculated the acceleration down the ramp. What is the vertical component of that?

3. Feb 21, 2016

### muhammed_oli

Hm alright, so the range equation I used was for symmetric projectile motion. I get what you are saying, I seem to mess up once I start using anything beyond the basics.

Okay for C I found the components for final ramp velocity to be 1.56i - 1.85j
v=d/t
Vf=d/t
find t
y=V(o)t-g/2(t^2)
for t? I hate this equation, is there anything easier? I tried a=v/t but wasnt sure of final velocity. edit i think im barking up the wrong tree with this now.

part d
ramp
5.75m/s^2(y direction)
a=v/t
v=1.85-0
5.75=1.85(t)
t=5.75/1.85

4. Feb 21, 2016

### haruspex

That's the one you need.
You could not use v=d/t even if you knew the final speed. Do you know why?

For d, How did you get 5.75 for the vertical component of acceleration down the ramp? (It should be less.)
a=v/t is fine, but that does not seem to be what you did.

5. Feb 21, 2016

### muhammed_oli

part c
v=d/t is for constant velocity?

this is what im thinking, after i've found t from the other equation
x=v(o)t+(1/2)at^2
a=0
x=v(o)(m/s)t(s)
x=x(m)

part d
I used 7.51sin50 to get that answer. I'm not sure how to approach this.

Last edited: Feb 21, 2016
6. Feb 21, 2016

### haruspex

yes.

Yes.
Sorry, my mistake. 5.75 is right. But what you did next was wrong. Try that again.

7. Feb 22, 2016

### muhammed_oli

k, gonna solve for t(ramp) using kinematic first.
v(f) = v(o)+at
v(o) = 0
2.42 = 7.51t
t = 0.3222

same, find t for the fall
y = v(o)t-1/2(g)(t^2)
-2 = -1.85t-4.9t^2
4.9t^2+1.85t-2 = 0
t = 0.4774

t(tot) = 0.3222+0.4774 = 0.7996 sec

how does that look?

8. Feb 22, 2016

### haruspex

Looks about right, but you quote too many digits, given the inputs and less precise intermediate results.

9. Feb 22, 2016

### muhammed_oli

thank you beary much! I got it right, 3 sig digits