Projectile Kinetic Energy Calculation

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SUMMARY

The discussion focuses on calculating the change in kinetic energy (KE) for two balls involved in a collision on a ramp setup. The first ball (mass m1) is released from rest and impacts the second ball (mass m2) at the bottom of the ramp. The calculations utilize the conservation of energy principle to determine the final velocities of both balls, applying kinematics equations such as v=sqrt(2gh) and Δy = 0.5gt². The final kinetic energy is computed using KE = 0.5m1v1² + 0.5m2v2², confirming the approach is valid under the assumption of no air resistance and horizontal launch.

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Homework Statement



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.

Homework Equations



kinematics equations.

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 [itex]\Delta y = v_0t+\dfrac{1}{2}at^2[/itex]

The initial velocity in the y direction is zero.

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

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

[itex]\Delta x = v_0t \implies v_0 = \dfrac{x}{t} = \dfrac{x}{\sqrt{\dfrac{2y}{g}}}[/itex].

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 [itex]KE = 0.5m_1v_1^2 +0.5m_2v_2^2[/itex]
we can subtract these two and we are done.



Is this thinking correct? Or where could I have gone wrong?
 
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Assuming zero air resistance and horizontal launch etc - that would be correct.
Both balls should hit at the same time, at different horizontal distances.
 
oneplusone said:
Find the change in kinetic energy.

Then going back to [itex]KE = 0.5m_1v_1^2 +0.5m_2v_2^2[/itex]
we can subtract these two and we are done.
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?
 

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