Jerry has managed to hack into the operating software of Mechano (mM = 2.75 kg),

Question

Jerry has managed to hack into the operating software of Mechano (mM = 2.75 kg), the evil robot cat. The picture shows Jerry having Mechano fire a projectile (mp = 0.200 kg) at Tom, the cat.

Part A: When the cannon fires, Mechano rolls backward with a recoil speed of 0.630 m/s. What is the speed of the projectile (vp) when it emerges from the cannon? I have this part figured out. How do you find the answer to part B?

Part B: How much energy (?E) is released by the gunpowder that explodes inside the cannon when the projectile is fired?

Map d Jerry has managed to hack into the operating software of Mechano (mM = 2.75 kg), the evil robot cat. The picture shows Jerry having Mechano fire a projectile (mp0.200 kg) at Tom, the cat. Answer the two questions below using three significant figures. Part A: When the cannon fires, Mechano rolls backward with a recoil speed o 0.630 m/s. What is the speed of the projectile (va) when it emerges from the cannon? Number Vp =118.6625 |rm / s the projectile is fired? Number AE7.5

Explanation / Answer

a)
Apply conservation of momentum

momentum gained by projectile = momentum gained by canon in opposite direction.

mp*vp = mW*vW

vp = mW*vW/mp

= 2.75*0.63/0.2

= 8.6625 m/s

b) Energy released, delta_E = 0.5*mw*vw^2 + 0.5*mp*vp^2

= 0.5*2.75*0.63^2 + 0.5*0.2*8.6625^2

= 8.05 J

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