A 30-g bullet is fired with a horizontal velocity of 450 m/s andbecomes embedded in block B, which has a mass of 3 kg. After theimpact, block B slides on 30-kg carrier C until it impacts the endof the carrier. Knowing the impact between B and C is perfectlyplastic and the coefficient of kinetic friction between B and C is 0.2,determine (a) the velocity of the bullet and B after the first impact,(b) the final velocity of the carrier.

a)

b)

Explanation:

a) Let's use the conservation of linear momentum.

Where:

b) As we have an external force between B and C we can not use the conservation of linear momentum here. We need to use the work definition.

The work here is due to the friction force.

I hope it helps you!

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.