Linear Motion and Linear Momentum

[hairey94]

Homework Statement

A 12 g bullet shot vertically into a 5 kg block and lifting it upwards to a maximum height of 4 mm. The bullet traveled for 1 ms in the block before stopping completely. Calculate the depth of the bullet embedded in the block.

Relevant Equations

The answer should be 0.06 m but I can't get it.

Linear Motion Equation to get the common velocity of the block and bullet just after collision:
v²=u²+2gs, I set v=0 at max height of s = 0.004 m and g = -9.81 m s^-2
I got u = 0.28 m/s
Then I calculate the impulse of the block using formula J = mv - mu, where v=0 and u=0.28 m/s and I got J = 1.4 N s
The last attempt is to use the conservation of linear momentum since the collision is inelastic and kinetic energy is not conserved:
mv' = J where v' is the velocity of the bullet just before the collision
I got v' = 116.95 m/s and this value seems reasonable and I continue using linear motion equation:
s = v't + (1/2)gt² where the t = 0.001 s and g = -9.81 m s^-2
but at the end the answer for s is not 0.06 m. Please let me know where I did wrong.

 

[PeroK]

Homework Statement:: A 12 g bullet shot vertically into a 5 kg block and lifting it upwards to a maximum height of 4 mm. The bullet traveled for 1 ms in the block before stopping completely. Calculate the depth of the bullet embedded in the block.
Relevant Equations:: The answer should be 0.06 m but I can't get it.

Linear Motion Equation to get the common velocity of the block and bullet just after collision:
v²=u²+2gs, I set v=0 at max height of s = 0.004 m and g = -9.81 m s^-2
I got u = 0.28 m/s
Then I calculate the impulse of the block using formula J = mv - mu, where v=0 and u=0.28 m/s and I got J = 1.4 N s
The last attempt is to use the conservation of linear momentum since the collision is inelastic and kinetic energy is not conserved:
mv' = J where v' is the velocity of the bullet just before the collision
I got v' = 116.95 m/s and this value seems reasonable and I continue using linear motion equation:
s = v't + (1/2)gt² where the t = 0.001 s and g = -9.81 m s^-2
but at the end the answer for s is not 0.06 m. Please let me know where I did wrong.

What are you using to determine the depth the bullet travels into the block?

 

[hairey94]

What are you using to determine the depth the bullet travels into the block?

I used the linear motion equation:
s = ut + (1/2) at²

 

[PeroK]

I used the linear motion equation:
s = ut + (1/2) at²

How did you calculate $a$?

 

[haruspex]

where v' is the velocity of the bullet just before the collision

What does "just after the collision" mean here?
The collision is a process taking 1ms. During that time, how would you describe the motions of the bullet and the block?

 

[hairey94]

How did you calculate $a$?

a = -g = -9.81 m s^-2

 

[PeroK]

a = -g = -9.81 m s^-2

The deceleration of a bullet in a block is $g$??

 

[hairey94]

What does "just after the collision" mean here?
The collision is a process taking 1ms. During that time, how would you describe the motions of the bullet and the block?

That is the time for the bullet to stop inside the block = 1 ms not the collision time. During this 1 ms the bullet is already inside the block.

 

[hairey94]

The deceleration of a bullet in a block is $g$??

well you tell me if it is right or wrong... that is only my attempt that lead to the incorrect answer.

 

[PeroK]

well you tell me if it is right or wrong... that is only my attempt that lead to the incorrect answer.

$g$ is the acceleration due to gravity for an object in free-fall. If a bullet is fired into a block at high-speed, the deceleration has nothing to do with gravity!

 

[hairey94]

then please suggest a solution for this question

 

[Delta2]

i would like to point out a mistake (or two) i believe you did in the calculation of $v'=116.95$

First of all the bullet after the collision has momentum too , bullet moves also with velocity u=0.28m/s hence you should set $mv'=(M+m)u$ where M the mass of block and m the mass of bullet. But still i believe this is wrong, cause the way i interpret the problem, the collision lasts for $\Delta t=1ms$ and during this time we have the impulse of the weight of the block and of the bullet. So i believe the overall correct equation to find v' is $$mv'=(M+m)u+(M+m)g\Delta t$$

 

[PeroK]

then please suggest a solution for this question

The bullet traveled for 1 ms in the block before stopping completely.

You've calculated the speed of the bullet; you can get the deceleration from the time.

 

[hairey94]

i would like to point out a mistake (or two) i believe you did in the calculation of $v'=116.95$

First of all the bullet after the collision has momentum too , bullet moves also with velocity u=0.28m/s hence you should set $mv'=(M+m)u$ where M the mass of block and m the mass of bullet. But still i believe this is wrong, cause the way i interpret the problem, the collision lasts for $\Delta t=1ms$ and during this time we have the impulse of the weight of the block and of the bullet. So i believe the overall correct equation to find v' is $$mv'=(M+m)u+(M+m)g\Delta t$$

the suggested solution is more complicated than I am expected but thank you for your help. I've calculated using your equation and I got v' = 121.04 m s^-1 vs the old v' = 116.95 m s^-1
So what should I do from here?

 

[hairey94]

You've calculated the speed of the bullet; you can get the deceleration from the time.

Is that speed can be used for the motion of the bullet inside the block? My guess was speed of bullet before collision is the initial speed and the acceleration is -g. But you point out my acceleration is wrong which I can accept that, but how about the initial speed of the bullet?

 

[Delta2]

You can assume that the bullet decelerates with constant deceleration inside the block. SO you know that in $\Delta t=1ms$ the velocity goes from v' down to u, what equation would you use to find the deceleration?

 

[hairey94]

You can assume that the bullet decelerates with constant deceleration inside the block. SO you know that in $\Delta t=1ms$ the velocity goes from v' down to u, what equation would you use to find the deceleration?

can I just skip to use s = (1/2)(v'+0)$\Delta t$? Instead of finding the deceleration.

 

[Delta2]

can I just skip to use s = (1/2)(v'+0)$\Delta t$? Instead of finding the deceleration.

well yes.

 

[hairey94]

Ok thank you for your help. This question really took me hours to complete despite the underlying concepts involved are actually quite simple.

 

[Delta2]

A small correction that doesn't affect the end result a lot but still i should mention, the final velocity of the bullet is not 0 but it is u=0.28 so you should take $$s=\frac{1}{2}(v'+u)\Delta t$$.

 

[hairey94]

But the question stated that 1 ms before stopping. Should we just assume 0 instead of 0.28?

 

[PeroK]

But the question stated that 1 ms before stopping. Should we just assume 0 instead of 0.28?

The question is poor on that point. The bullet doesn't stop "completely", only relative to the block. You can get an approximation by neglecting this, but an accurate answer would not assume the bullet stops completely.

 

[hairey94]

ok really appreciated for the help. I think the question just want to make the calculation simpler but instead it is poorly worded that it becomes so much complicated as we tried to assume many things at once.

 

[haruspex]

That is the time for the bullet to stop inside the block = 1 ms not the collision time. During this 1 ms the bullet is already inside the block.

No, during that 1ms the block and bullet are traveling at different speeds. The collision is a process.

 

[ehild]

Homework Statement:: A 12 g bullet shot vertically into a 5 kg block and lifting it upwards to a maximum height of 4 mm. The bullet traveled for 1 ms in the block before stopping completely. Calculate the depth of the bullet embedded in the block.
Relevant Equations:: The answer should be 0.06 m but I can't get it.

Linear Motion Equation to get the common velocity of the block and bullet just after collision:
v²=u²+2gs, I set v=0 at max height of s = 0.004 m and g = -9.81 m s^-2
I got u = 0.28 m/s

The equation holds for motion with constant acceleration, but the block does not move so. In the first stage of its motion, the bullet enters it and moves inside it, making the block accelerate, while it decelerates, till they both have the same velocity, V, That is the end of the collision process, and it happens in 1 ms. During that time, the block rises by s1.,
In the second stage, the block (with the bullet inside) rises with decreasing speed, with constant deceleraion of g, and covers the distance s2.
The total distance traveled is s1+s2= 4 mm.
In the first stage, the initial velocity of the block is zero, the final one is V and the time is 1 ms. You can determine s1 with these data.
In the second stage, the initial velocity of the block is V, the final one is zero, the acceleration is -g. You can calculate the distance from these data. The equation s1+s2= 4mm has one unknown V, the velocity after the collision. Determine it.
When you know V, the velocity after collision, you can proceed as you planned to calculate the initial velocity of the bullet.
The distance traveled by the bullet inside the block can be determined with its relative speed inside the block and the time duration of the collision process.

 

[LordJulius]

the suggested solution is more complicated than I am expected but thank you for your help. I've calculated using your equation and I got v' = 121.04 m s^-1 vs the old v' = 116.95 m s^-1
So what should I do from here?

can you explain how you get 121.04??

 

[haruspex]

can you explain how you get 121.04??

The thread is over a year old. Hairey94 doesn’t seem to be around any more.
If you have the same homework problem, please post your attempt.

 

[LordJulius]

i would like to point out a mistake (or two) i believe you did in the calculation of $v'=116.95$

First of all the bullet after the collision has momentum too , bullet moves also with velocity u=0.28m/s hence you should set $mv'=(M+m)u$ where M the mass of block and m the mass of bullet. But still i believe this is wrong, cause the way i interpret the problem, the collision lasts for $\Delta t=1ms$ and during this time we have the impulse of the weight of the block and of the bullet. So i believe the overall correct equation to find v' is $$mv'=(M+m)u+(M+m)g\Delta t$$

can you explain how you get the formula in more detail ?? I'm abit confuse on how you get that formula

 

[LordJulius]

can you explain how you get 121.04??

no needdd i already found the prooblem on why i can't get the answer

 

[Delta2]

can you explain how you get the formula in more detail ?? I'm abit confuse on how you get that formula

I am using the Impulse-Momentum Theorem (which is a direct corollary from Newton's second law). Hold on while I 'll write more details about this.