Question

Find the force that they exert on each other.

Answer 1

https://serc.carleton.edu/dmvideos/players/cart_push_off_2_AD.html

Answer 2

@sillybilly123 When you're online :)

Answer 3

I'd probably wait for silly to confirm

Answer 4

actually it should probably be vf^2 = v_0^2 + 2a*delta(x) instead since the first kinematics equation is less useful in this situation

Answer 5

idk what I'm doing just wait for silly

Answer 6

Yeah. Due to the different masses, you get different speeds. Due to Newton's Third Law though, the force should be the same (that they are exerting on each other). But you end up with different accelerations, due to the differing speeds. And thus, different forces....

Answer 7

They both start at rest, so a different speed is a different acceleration.

Answer 8

Yeah, I am curious as to how to go about doing this one.

Answer 9

You can looking at Impulse ..... \( F \cdot \Delta t\), specifically:

\(F \Delta t = m \Delta v \implies F_{ave} = m \dfrac{\Delta v}{\Delta t}\)


This is a strangulation of the second law, \(F = m \dfrac{dv}{dt}\), the idea being to calculate **average force**, \(F\), over the contact time \(\Delta t\). Note that acceleration during the push does not have to be constant, and most likely never is.

On the LHS, \(m_1 \approx 148.4 kg\), on the RHS, \(m_2 \approx 91.4 kg\)

Contact lasts **approximately** from frame f = -188 to f = - 30

At 240 fps, that's \(\Delta t = \dfrac{-30 - (-188)}{240} \approx 0.66 s \)

As soon as contact is lost, both carts are decelerating due to friction etc. So that 20cm trip gives you only an estimate of final velocity.


On LHS the 20cm trip starts at frame f = 0 and ends at f = 68. I make that \(v_1 \approx 0.7 m/s\)

On RHS same calculation gives \(v_2 \approx 1.1 m/s\)

So we can so the calculations:

\(F_1 = 148.4 \dfrac{0.7}{0.66} = 157.4 ~ N\)

\(F_2 = 91.4 \dfrac{1.1}{0.66} = 152.3 ~ N\)

In theory they should be the same but they are close enough to suggest there are rounding and observation errors (fps)

That's **how**.

This is the same as calculating **average** acceleration during the contact phase for each cart, and using \(F = m a_{ave}\), thing as \(a_{ave} = \dfrac{\Delta v}{\Delta t}\)

Passes a reality check too (compare it say to body weight) but be careful with the numbers and calcs and rounding and all that stuff........