Solving Impulse: Why Only 1 Method Works for Determining Change in Velocity

[Nito]

Homework Statement

A ball of mass 0.150 kg is dropped from rest from a
height of 1.25 m. It rebounds from the floor to reach a
height of 0.960 m. What impulse was given to the ball
by the floor?

Homework Equations

Δmv/t =I
vf^2=vo^2 +2aΔy

The Attempt at a Solution

mghi = mghf +1/2mvf^2

vf = 2.38m/s

0.150kg x (2.38-0) = 0.35 Kg m/s

But this did not work. What the answer key did was use kinematics equations to find the velocity before the ball hit the ground and then found the velocity when the ball reached the height 0.960m. They then did change in velocity x mass to find the impulse.

Essentially, I don't see why using the conservation of energy equation to find the change in velocity is wrong. Why does this not work?

 

[Nito]

nvm, there's no kinetic energy when it reaches its peak height. stupid lol!

 

[Chestermiller]

Homework Statement

A ball of mass 0.150 kg is dropped from rest from a
height of 1.25 m. It rebounds from the floor to reach a
height of 0.960 m. What impulse was given to the ball
by the floor?

Homework Equations

Δmv/t =I
vf^2=vo^2 +2aΔy

The Attempt at a Solution

mghi = mghf +1/2mvf^2

vf = 2.38m/s

0.150kg x (2.38-0) = 0.35 Kg m/s

But this did not work. What the answer key did was use kinematics equations to find the velocity before the ball hit the ground and then found the velocity when the ball reached the height 0.960m. They then did change in velocity x mass to find the impulse.

Essentially, I don't see why using the conservation of energy equation to find the change in velocity is wrong. Why does this not work?

Because mechanical energy is not conserved in this collision. Part of the mechanical energy is converted to thermal energy (heat) in the collision, with the temperature of the ball and the temperature of the ground increasing slightly.

 

[pgardn]

Homework Statement

A ball of mass 0.150 kg is dropped from rest from a
height of 1.25 m. It rebounds from the floor to reach a
height of 0.960 m. What impulse was given to the ball
by the floor?

Homework Equations

Δmv/t =I
vf^2=vo^2 +2aΔy

The Attempt at a Solution

mghi = mghf +1/2mvf^2

vf = 2.38m/s

0.150kg x (2.38-0) = 0.35 Kg m/s

But this did not work. What the answer key did was use kinematics equations to find the velocity before the ball hit the ground and then found the velocity when the ball reached the height 0.960m. They then did change in velocity x mass to find the impulse.

Essentially, I don't see why using the conservation of energy equation to find the change in velocity is wrong. Why does this not work?

Like chester said, the ball started off with a certain amount of potential energy mg(1.25) and then ended with mg(0.96). So you would have appeared to "lose" energy.

Often if time is involved in a problem, it will be difficult to use the conservation of mechanical energy alone to solve that problem. If you can use kinematics, cons. of E, and cons. of momentum, you are good on a significant number of mechanics problems. And being able to draw free body diagrams...