To view the video above, you?ll need to have a Quicktime player installed on you

Question

To view the video above, you?ll need to have a Quicktime player installed on your computer. Use the controls at the bottom of the video. Use the right and left arrows to advance the video frame-by-frame. The videos shows a peanut M&M; candy fired out of a pneumatic cannon into a foam block attached to a cart on a low-friction track. The M&M; sticks into the foam and remains there after the collision. In this case, the speed of the M&M; is too high to measure directly from the video. Instead, we?ll use measurements of the speed of the cart after the collision to determine the speed of the M&M;. The mass of the M&M; candy launched out of the cannon is 2.37 grams. The combined mass of the cart and foam block is 0.393 kg. How many frames does it take for the cart to move 15cm after the collision? (Do not enter units for this answer.) frames What is the speed of the cart as it moves 15 cm after being struck by the M&M; candy? Assuming that momentum is conserved during the collision, what must be the speed of the M&M; before it hits the cart? Use the video to estimate the time for the collision. That is, estimate how long it takes for the moving M&M; to become embedded in the foam block. Note that your estimate will not be very precise, because the time is very short. Based on the measurements you?ve made, what is the magnitude of the average force the M&M; applies to the cart during the collision?

Explanation / Answer

(3) Assuming that momentum is conserved during collision, the speed of the M&M before it hits the cart which is given as :

using conservation of momentum, we have

pbefore = pafter

m1 v1 = m2 v2

where, m1 = mass of the M&M candy = 2.37 g = 0.00237 kg

m2 = combined mass of the cart and foam block = 0.393 kg

v2 = speed of the cart = 0.379 m/s

inserting the values in above relation,

(0.00237 kg) v1 = (0.393 kg) (0.379 m/s)

v1 = (0.148947 kg.m/s) / (0.00237 kg)

v1 = 62.8 m/s

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