Projectile Kinetic Energy Calculation

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In summary, the conversation discusses a scenario where two balls, one at rest on a ramp and one at the top of a table, collide and then fall to the floor. The distance traveled by each ball, as well as the height of the table and ramp, are recorded. The goal is to find the change in kinetic energy of the system, using equations for conservation of energy and kinematics. The conversation also mentions that the scenario assumes no air resistance and horizontal launch. The final step is to calculate the initial velocities of both balls immediately after impact, and then use the equation for kinetic energy to find the change in KE.
  • #1
oneplusone
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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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  • #2
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
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?
 

FAQ: Projectile Kinetic Energy Calculation

What factors affect the distance a projectile can be thrown?

The distance a projectile can be thrown is primarily affected by the initial velocity, angle of launch, and air resistance. Other factors such as wind, temperature, and altitude may also play a role.

How does the mass of a projectile affect its trajectory?

The mass of a projectile does not affect its trajectory as long as the initial velocity and angle of launch remain constant. However, a heavier projectile will require more force to be thrown the same distance as a lighter projectile due to its greater inertia.

What is the ideal angle to launch a projectile for maximum distance?

The ideal angle to launch a projectile for maximum distance is 45 degrees. This angle allows for the greatest combination of horizontal and vertical velocity, resulting in the longest possible range.

How does air resistance affect the trajectory of a projectile?

Air resistance, also known as drag, slows down the forward motion of a projectile and causes it to follow a curved path. The amount of air resistance depends on the shape and surface area of the projectile, as well as the density of the air it is moving through.

Can a projectile be thrown in a perfect parabolic arc?

In a vacuum with no air resistance, a projectile can be thrown in a perfect parabolic arc. However, in the real world with air resistance, the trajectory will be slightly affected and the arc will not be perfectly symmetrical.

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