# Speed of block sliding on frictionless ramps

Here's the question:

My book says the answer is C. How is it not A?

I know that all the potential energy is transferred to kinetic energy. With algebra, knowing Kinetic energy is (1/2) * m * v^2 and gravitational potential is mgh, I solve for h which results in (v^2)/2g

Ok so since this is a proportional reasoning problem my focus is that h = v^2 meaning height is directly proportional to the square of velocity. That being said, if we half the velocity, that means that some value (let's use p for the variable) multiplied by height gets me to (1/2v)^2.

That being said, if we refer back to the equation h = (v)^2 and v is being halved, it should look like this:

h * p (<- growth factor) = (0.5v)^2. Now it looks like the right side of the equation grew by a factor of (1/4). Think about it. If your velocity is 10, the right side of the equation becomes 100. If you half that velocity, the right side of that equation becomes 25. You can see that the right side of the equation grew by a factor of 1/4. That means that $p$ (my growth factor variable) should also be (1/4). $h * 1/4 = h/4$. NOT $3h/4$. Where did I go wrong?

• I believe I see part of my error. The question said it continues to slide up another smooth ramp. But still, how would I go about solving this? – Maheer Aeron Feb 7 '17 at 1:43

The potential energy at the start is $mgh$, which is converted into kinetic energy of $\frac 12mv^2$ at the bottom. When the velocity is $\frac v2$ the kinetic energy is $\frac 12m(\frac v2)^2=\frac 18mv^2$. As $\frac 14$ of the energy is now kinetic, the rest must be potential, so the potential energy is $\frac 34mgh$ the the height is $\frac 34h$

I got lost in your explanation, so can't help you figuring out where you went wrong. Your attempt to solve it with conservation of energy is certainly the easiest and correct.

Let's analyze the question

• smooth ramp: this is meant to mean no friction, no energy is lost
• starts from rest at height $h$: at this point there is only potential energy ("rest") and it is equal to $mgh$
• at the bottom it is moving at speed $v$: at this point all potential energy has been converted into kinetic enrgy equal to $mv^2/2$
• slides up a second ramp. at what height $h^*$ is the speed equal to $v/2$? At this point the block has both potential energy equal to $mgh^*$ and kinetic energy equal to $m(v/2)^2/2$

Because of the smoothness of the ramps energy is conserved and you can put the energies at all three points (start, bottom of ramp and at height $h^*$) equal. This gives:

$$mgh=\frac{m}{2}v^2=mgh^*+\frac{m}{2}\left(\frac{v}{2}\right)^2$$

Eliminating $v$ from these equations you end up with

$$h^*=\frac{3}{4}h$$

I have another solution:

Ok basically we can think of 2 phases. One where it goes from the first ramp down. The other phase is from down to another ramp, however this time at some height where velocity is equal to half the velocity at the bottom.

Considering the first phase, we know that all potential get's transformed into kinetic. Thus: 1/2mv^2 = mgh. Solving for v (but leaving it squared), we get v^2 = 2gh. I'm going to leave that for now, and as a matter of fact, I will call that v1 so (v1)^2 = 2gh.

Now for the second phase, at the bottom it has the same velocity as the final velocity at the first phase. All this energy gets transferred to a mix of some kinetic and some potential (and this is because we aren't going all the way back up to the top where velocity is 0 again).

Therfore, (1/2)m(v1)^2 = (1/2)m(v2)^2 + mgH.

Now, I used capital H because this is the actual height I am looking for. v2 is the velocity that we must reference to find capital H. Before I move on to what that equals, let me divide out mass.

(1/2)(v1)^2 = (1/2)(v2)^2 + gH.

Ok so the question is saying to find H when we are at half the velocity at the bottom. so v2 = (1/2v1). Therefore:

(1/2)(v1)^2 = (1/2)((1/2)(v1))^2 + gH

Now for (1/2)((1/2)(v1))^2, I distribute out the ^2. so (1/2)(1/4)(v1)^2 which simplifies to (1/8)(v1)^2. Now back to the equation:

(1/2)(v1)^2 = (1/8)(v1)^2 + gH

Combining like terms I get this now:

(3/8)(v1)^2 = gH.

Looks like we are getting somewhere. Remember that (v1)^2 = 2gh? I'm going to substitute that for (v1)^2.

(3/8)(2gh) = gH

divide g out

(3/8)(2h) = H

Multiply the 2

(3/4)h = H or H = (3h/4).