Speed of Candy - Data and Calculations The mass of the M&M;? candy launched out

Question

Speed of Candy - Data and Calculations The mass of the M&M;? candy launched out of the cannon is 2.23 grams. The combined mass of the cart and foam block is 0.453 kg. How many frames does it take for the cart to move 15 cm after the collision? (Do not enter units for this answer.) 63 frames What is the speed of the cart as it moves 15 cm after being struck by the M&M;? candy? .5769 m/s Assuming that momentum is conserved during the collision, what must be the speed of the M&M;? before it hits the cart? Average Force Exerted by Candy - Data and Calculations 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. .0208 s 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

a few details are missing . it would help if there was a proper diagram . But i am going to imagine it in the most simple and convinient way , So stay with me .

the mass of the projectile is 2.23 gm = 0.00223 kg .

it is launched into a cart containing foam and their total mass is 0.453 kg .

Now , it was calculated that the speed of ( candy + cart + foam ) = 0.5769 m/s .

So , the momentum after collision is total weight of  ( candy + cart + foam ) * velocity of this mass = (0.00223+0.453)*0.5769 = 0.262622 kgm/s .

If the momentum is conserved , the total momentum of the system should be same even before the collision . Now , assuming the cart + foam is initaially at rest , their momentum = 0 & momentum of candy is given by (its mass*earlier velocity) which must be equal to 0.262622 kgm/s . So , 0.00223*v=0.262622 . So , v = 117.77 m/s .

Now , for next part , the andy , starts with u = 117.77m/s and ends with v = 0.5769m/s in a time of 0.0208s . So , the acceleration is (v-u)/t = (0.5769-117.77)/0.0208 = 5634.1737 m/s2 . This multiplied by its mass , gives the average force applied on the cart . So , the required answer is 5634.1737*0.00223 = 12.5642N .

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