What Is the Block's Velocity After the Projectile Emerges?

[Mesmerr]

Homework Statement

a 200-g projectile is fired with a velocity of 900m/s towards the center of the 15-kg wooden block, which rests on a rough surface. If the projectile penetrates and emerges from the block with a velocity of 300m/s, determine the velocity of the block just after the projectile emerges. How long does the block slide on the rough surface, after the projectile emerges, before it comes to rest again? the coefficient of kinetic friction between the surface and the clock is μk=0.2.

Homework Equations

conservation of momentum

∫ƩFdt=∫mdv

Fr=N*μk

N=W

W=m*g

and maybe ƩF=ma?

The Attempt at a Solution

Attempt is an attached pdf. I am stuck on how to find the force the bullet applied to the block. With that I could do Fb-Fr=ƩF, ƩF(t2-t1)=∫mdv where Δt would be my answer. Any guidance would be appreciated. Thank you!

 

[SammyS]

Homework Statement

a 200-g projectile is fired with a velocity of 900m/s towards the center of the 15-kg wooden block, which rests on a rough surface. If the projectile penetrates and emerges from the block with a velocity of 300m/s, determine the velocity of the block just after the projectile emerges. How long does the block slide on the rough surface, after the projectile emerges, before it comes to rest again? the coefficient of kinetic friction between the surface and the clock is μk=0.2.

Homework Equations

conservation of momentum

∫ƩFdt=∫mdv

Fr=N*μk

N=W

W=m*g

and maybe ƩF=ma?

The Attempt at a Solution

Attempt is an attached pdf. I am stuck on how to find the force the bullet applied to the block. With that I could do Fb-Fr=ƩF, ƩF(t2-t1)=∫mdv where Δt would be my answer. Any guidance would be appreciated. Thank you!

Hello Mesmerr. Welcome to PF !


For what length of time is the bullet passing through the block? You can answer this if you assume that the bullet experiences the same force throughout its path through the block.

 

[Mesmerr]

The question was how long does the block slide on the surface after the bullet leaves it. I noticed though that it said "after the bullet leaves the block". This means that the bullet is no longer acting on the block meaning you can use the equation of impulse to give you.

ƩF_blockΔt=m_blockΔv where ƩF = F_friction

Thanks for the help though! I do see how if you are given the time through the block you can get impulse on the bullet. Then because of Newton's laws the block would experience the same force.

 

[cepheid]

Why can't you just use conservation of momentum (for the bullet-block system) to solve this?

EDIT: I see. Because external forces (friction) act on the system. OK. I don't see a problem with SammyS's approach. Using the impulse momentum theorem, you can at least figure out how much force the bullet applies to the block.