A 0.05 kg weight is used to accelerate a cart. At t=1 second, the 0.5 kg cart is at x=0.5 m, at rest. At t=2.5 seconds, the cart is at 1.5 m, traveling at 1.3 ± 0.3 m/s. Following the procedure for the first part of the experiment, compute and compare the impulse and change in momentum.

Question

Physics

Sophia Hess

2 August, 18:08

A 0.05 kg weight is used to accelerate a cart. At t=1 second, the 0.5 kg cart is at x=0.5 m, at rest. At t=2.5 seconds, the cart is at 1.5 m, traveling at 1.3 ± 0.3 m/s. Following the procedure for the first part of the experiment, compute and compare the impulse and change in momentum.

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Jaxon · Answer 1

1) Change in momentum, Δp

Δp = p2 - p1 = mV2 - mV1 = m (V2 - V1) = 0.5 kg * (1.3 m/s - 0) = 0.65 N*s

2) Impulse = F*Δt

F = m*a

Vf^2 = Vo^2 + 2aΔx = > a = [Vf^2 - Vo^2] / (2Δx)

a = [ (1.3 m/s) ^2 - 0] / [2 (1.5m - 0.5m) ] = (1.3m/s) ^2 / 2m = 0.845 m/s^2

F = m*a = 0.5kg * 0.845 m/s^2 = 0.4225 N

I = F*Δt = 0.4225N * (2.5 s - 1s) = 0.63375 N*s

Conclusion:

Δp = 0.65 N*s

I = 0.63 N*s

Which are reasonably equal.

You have to take into account the experimental errors, which for the velocity is + / - 0.3 m/s.

That error is what makes the difference between the impulse and the change in momentum.

You can find the range of error by using the maximum and the minimum velocities of the range, i. e. 1.6 m/s and 1.0 m/s, and then you will find that the the Impulse and the Change of momentum are the same, inside the range of error provided by of the measurements.