Homework Help: Throwing a projectile

1. Nov 14, 2013

oneplusone

1. The problem statement, all variables and given/known data

There's a ramp at the top of a table. You hold a ball of mass m1 at the top of the ramp, and release it starting at rest, so it hits ball with mass m2 which is initially at rest at the bottom of the ramp.

The ramp is set up, so the first ball hits the second, and they fall off the table and hit the floor. Also, when they fall off the ramp, it is parallel to the ground (so basically there's no y component of velocity).

You record the distances both balls travel, the height of the table, and height of the ramp.

Find the change in kinetic energy.

2. Relevant equations

kinematics equations.

3. The attempt at a solution

This is more of a lab, so KE is not conserved (lost in sound).

First I used conservation of energy to find the final velocity of the balling falling. that would just be v=sqrt(2gh)
The total kinetic energy would be then: 1/2 (m1)(2gh)

Next,

Using $\Delta y = v_0t+\dfrac{1}{2}at^2$

The initial velocity in the y direction is zero.

So we have: $\Delta y = 0.5gt^2 \implies t = \sqrt{\dfrac{2y}{g}}$.

Now from the x component, we have the same equation , but there is no acceleration so we have:

$\Delta x = v_0t \implies v_0 = \dfrac{x}{t} = \dfrac{x}{\sqrt{\dfrac{2y}{g}}}$.

Now plugging in the values collected for y and x, we can calculate the "initial velocities' of both masses (immediately after impact).

Then going back to $KE = 0.5m_1v_1^2 +0.5m_2v_2^2$
we can subtract these two and we are done.

Is this thinking correct? Or where could I have gone wrong?

2. Nov 14, 2013

Simon Bridge

Assuming zero air resistance and horizontal launch etc - that would be correct.
Both balls should hit at the same time, at different horizontal distances.

3. Nov 14, 2013

haruspex

If v1 and v2 refer to the velocities just after impact, you have correctly found the change in KE from just before the collision to just after. Or are these the velocities at floor level?