Question

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.

Answer 1

a)
`v_(fb+B)=4.46\: m/s`

b)
`v_(fc)=1.21\: m/s`

Step-by-step explanation:

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

`m_(b)v_(iv)=m_(B)v_(f)+m_(b)v_(f)`

Where:

• m(b) is mass of bullet (0.03 kg)
• v(ib) is the initial velocity of the bullet (450 m/s)
• m(B) is the mass of the block (3 kg)
• v(f) is the final velocity of the bullet and the block

`m_(b)v_(iv)=m_(B)v_(f)+m_(b)v_(f)`

`v_(f)=(m_(b)v_(iv))/(m_(B)+m_(b))`

`v_(f)=(0.03*450)/(0.03+3)`

`v_(fb+B)=4.46\: m/s`

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.

`W=\Delta K`

`W=K_(f)-K_(i)`

`-\mu m_(B)g=0.5(m_(b)+m_(B)+m_(C))v_(f)^(2)-0.5(m_(b)+m_(B))v_(i)^(2)`

`-\mu m_(B)g=0.5(m_(b)+m_(B)+m_(C))v_(f)^(2)-0.5(m_(b)+m_(B))v_(i)^(2)`

`-0.2*3*9.81=0.5(0.03+3+30)v_(f)^(2)-0.5(0.03+3)4.46^(2)`

`v_(f)^(2)=(-0.2*3*9.81+0.5(0.03+3)4.46^(2))/(0.5(0.03+3+30))`

`v_(fC)=1.21\: m/s`

I hope it helps you!