Bullet shooting vertically into a block

[christang_1023]

Homework Statement

A 28-g bullet is shot vertically into an 6-kg block. The block lifts upward 5mm (see the figure). The bullet penetrates the block and comes to rest in it in a time interval of 0.0010s. Assume the force on the bullet is constant during penetration and that air resistance is negligible. The initial kinetic energy of the bullet is closest to
A.63J
B.32J
C.0.29J
D.1.4×10^(-3)J
E.42J

Homework Equations

According to conservation of energy, I get: E0=ΔE+mgh, where E0 is initial kinetic energy; ΔE is the energy lost during the penetration; m is the mass of the bullet; h is the height of the block.

The Attempt at a Solution

However, I fail to get ΔE, because of unknown F (the constant force on the bullet exerted by the block)

 

[neilparker62]

Use $$ΔE = ½ \frac { m_1m_2 } { m_1+m_2 } \timesΔv^2 $$

 

[christang_1023]

Use $$ΔE = ½ \frac { m_1m_2 } { m_1+m_2 } \timesΔv^2 $$

Thank you for your reply.
I try using your formula to calculate ΔE, regarding Δv equal to the velocity after the bullet has risen by 5mm, but only to get answer D. However, the corret answer should be A. 63J.
I feel confused.

 

[ehild]

Homework Statement

A 28-g bullet is shot vertically into an 6-kg block. The block lifts upward 5mm (see the figure). The bullet penetrates the block and comes to rest in it in a time interval of 0.0010s. Assume the force on the bullet is constant during penetration and that air resistance is negligible. The initial kinetic energy of the bullet is closest to
A.63J
B.32J
C.0.29J
D.1.4×10^(-3)J
E.42J
View attachment 234612

Homework Equations

According to conservation of energy, I get: E0=ΔE+mgh, where E0 is initial kinetic energy; ΔE is the energy lost during the penetration; m is the mass of the bullet; h is the height of the block.

h is not the height of the block, but the height the block rises after the collision with the bullet.

The Attempt at a Solution

However, I fail to get ΔE, because of unknown F (the constant force on the bullet exerted by the block)

You do not need the force acting on the bullet. Knowing the height, and using conservation of energy, you get the initial speed of the block with the bullet inside just after the bullet penetrated into the block.
Knowing this speed, use conservation of momentum to get the speed of the bullet before the collision.

 

[neilparker62]

I get answer A using the formula you started with and ΔE as given

 

[PeroK]

Use $$ΔE = ½ \frac { m_1m_2 } { m_1+m_2 } \timesΔv^2 $$

I think at this level, the OP should be trying to work with momentum and energy, rather than picking up a formula like this and just plugging in some numbers.

Thank you for your reply.
I try using your formula to calculate ΔE, regarding Δv equal to the velocity after the bullet has risen by 5mm, but only to get answer D. However, the corret answer should be A. 63J.
I feel confused.

First, energy is not conserved. For energy to be conserved, the bullet would have to bounce elastically off the block. What you have is an inelastic collision, where energy is lost.

However, even in an inelastic collision, you have conservation of momentum: in a closed system, you always have conservation of momentum.

What can you do with conservation of momentum?

PS I see @ehild has already said this!

 

[neilparker62]

I think at this level, the OP should be trying to work with momentum and energy, rather than picking up a formula like this and just plugging in some numbers!

Agree 100% - one should never just plug numbers into a formula without understanding where the formula comes from. But I think my advice to the OP should rather be to derive the formula for himself/herself using momentum and energy as you say. But thereafter use it without apology. No point continually reinventing the wheel!

 

[PeroK]

Agree 100% - one should never just plug numbers into a formula without understanding where the formula comes from. But I think my advice to the OP should rather be to derive the formula for himself/herself using momentum and energy as you say. But thereafter use it without apology. No point continually reinventing the wheel!

I like inventing things. In this case, for example, I invented:

$\Delta E = \frac{M^2gh}{m} \ \ (m << M)$

 

[christang_1023]

Thank you for your help, and I've figured it out! :)

 

[neilparker62]

I like inventing things. In this case, for example, I invented:

$\Delta E = \frac{M^2gh}{m} \ \ (m << M)$

For the record neither my formula nor yours are "invented" since both have their origin here (and no doubt elsewhere similarly!):

http://hyperphysics.phy-astr.gsu.edu/hbase/inecol.html#c2

But what matters is the OP obtained clarity from ehild's explanation.