Collision/Impulse: Calculating Impulse and Average Force

[blue5t1053]

I'm unsure if I set this up correctly. It's a mulitple choice homework question and it doesn't match any possible answers.

Question:
A 0.605 kg ball drops vertically onto a floor, hitting with a speed of 34 m/s. It rebounds with an initial speed of 11 m/s. (a) What impulse acts on the ball during the contact? (b) If the ball is in contact with the floor for 0.0171 s, what is the magnitude of the average force on the floor from the ball?

Attempt:
Part A:
I = mvf - mvi
I = m(vf - vi)
I = 0.605(11-34)
I = -13.915 kg-m/sec

Attempt:
Part B:
If the contact time = 0.0171 s, then:
F = Δp/Δt = I/Δt = -13.915 kg-m/sec/0.0171 s
F = -813.743 Nt

 

[Doc Al]

Attempt:
Part A:
I = mvf - mvi
I = m(vf - vi)
I = 0.605(11-34)
I = -13.915 kg-m/sec

Incorrect: Realize that momentum is a vector. Direction (sign) counts! (Call up + and down -.)

Attempt:
Part B:
If the contact time = 0.0171 s, then:
F = Δp/Δt = I/Δt = -13.915 kg-m/sec/0.0171 s
F = -813.743 Nt

Right idea, but you need to fix part A first. Also: magnitudes are always positive.

 

[blue5t1053]

Incorrect: Realize that momentum is a vector. Direction (sign) counts! (Call up + and down -.)


Right idea, but you need to fix part A first. Also: magnitudes are always positive.

You're right about the magnitude! Thanks, I forgot that.

So I have...

Re-Attempt:
Part A:
I = mvf - mvi
I = m(vf - vi)
I = 0.605(11-(-34))
I = 27.225 kg-m/sec

Re-Attempt:
Part B:
If the contact time = 0.0171 s, then:
F = Δp/Δt = I/Δt = 27.225 kg-m/sec/0.0171 s
F = 1592.11 Nt

 

[Doc Al]

Looks good! (The abbreviation for Newtons is just N, not Nt.)