Linear Motion and Linear Momentum

In summary, the problem involves a bullet with a mass of 12 g being shot into a 5 kg block, lifting it upwards to a maximum height of 4 mm. The bullet travels for 1 ms before stopping completely. The goal is to calculate the depth of the bullet embedded in the block, with the correct answer being 0.06 m. To solve this, the linear motion equation is used, with the deceleration (a) being set to -9.81 m s-2. However, this leads to an incorrect answer. Another attempt is made using the conservation of linear momentum, taking into account the momentum of the bullet before and after collision. This leads to a velocity of 116.95 m/s
  • #1
hairey94
15
2
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.
 
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  • #2
hairey94 said:
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?
 
  • #3
PeroK said:
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
 
  • #4
hairey94 said:
I used the linear motion equation:
s = ut + (1/2) at2
How did you calculate ##a##?
 
  • #5
hairey94 said:
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?
 
  • #6
PeroK said:
How did you calculate ##a##?
a = -g = -9.81 m s-2
 
  • #7
hairey94 said:
a = -g = -9.81 m s-2
The deceleration of a bullet in a block is ##g##??
 
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  • #8
haruspex said:
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.
 
  • #9
PeroK said:
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.
 
  • #10
hairey94 said:
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!
 
  • #11
then please suggest a solution for this question
 
  • #12
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 $$mv'=(M+m)u+(M+m)g\Delta t$$
 
  • #13
hairey94 said:
then please suggest a solution for this question

hairey94 said:
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.
 
  • #14
Delta2 said:
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?
 
  • #15
PeroK said:
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?
 
  • #16
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?
 
  • #17
Delta2 said:
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.
 
  • #18
hairey94 said:
can I just skip to use s = (1/2)(v'+0)##\Delta t##? Instead of finding the deceleration.
well yes.
 
  • #19
Ok thank you for your help. This question really took me hours to complete despite the underlying concepts involved are actually quite simple.
 
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  • #20
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 $$s=\frac{1}{2}(v'+u)\Delta t$$.
 
  • #21
But the question stated that 1 ms before stopping. Should we just assume 0 instead of 0.28?
 
  • #22
hairey94 said:
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.
 
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  • #23
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.
 
  • #24
hairey94 said:
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.
 
  • #25
hairey94 said:
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.
 
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  • #26
hairey94 said:
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??
 
  • #27
LordJulius said:
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.
 
  • #28
Delta2 said:
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
 
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  • #29
LordJulius said:
can you explain how you get 121.04??
no needdd i already found the prooblem on why i can't get the answer
 
  • #30
LordJulius said:
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.
 
  • #31
According to the Impulse-Momentum Theorem the change in the momentum of a system of bodies is equal to the total impulse of the external forces acting on the system.
So:
Change in momentum of the system bullet+block ##\Delta P=(M+m)u-mv##
Impulse of external forces acting on the system: The external forces are the weight of the bullet ##mg## and the weight of the block ##Mg##. Their impulses , by definition, are equal to ##-mg\Delta t## and ##-Mg\Delta t## (we take as positive the up direction, that is the direction of the initial velocity of the bullet ##v##).
So by impulse-momentum theorem we have $$(M+m)u-mv=-mg\Delta t-Mg\Delta t$$ and the final equation follows with just a tiny bit of algebra.
 
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  • #32
Delta2 said:
According to the Impulse-Momentum Theorem the change in the momentum of a system of bodies is equal to the total impulse of the external forces acting on the system.
So:
Change in momentum of the system bullet+block ##\Delta P=(M+m)u-mv##
Impulse of external forces acting on the system: The external forces are the weight of the bullet ##mg## and the weight of the block ##Mg##. Their impulses , by definition, are equal to ##-mg\Delta t## and ##-Mg\Delta t## (we take as positive the up direction, that is the direction of the initial velocity of the bullet ##v##).
So by impulse-momentum theorem we have $$(M+m)u-mv=-mg\Delta t-Mg\Delta t$$ and the final equation follows with just a tiny bit of algebra.
okayy i understand it now thank you so much
 
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Related to Linear Motion and Linear Momentum

1. What is linear motion?

Linear motion refers to the movement of an object in a straight line, with a constant speed and direction. This type of motion is also known as rectilinear motion.

2. What is linear momentum?

Linear momentum is the product of an object's mass and velocity. It is a measure of the quantity of motion an object possesses in a certain direction.

3. How is linear momentum conserved?

According to the law of conservation of momentum, the total linear momentum of a closed system remains constant, meaning it is conserved. This means that in a closed system, the total momentum before and after a collision or interaction remains the same.

4. What is the difference between linear and angular momentum?

Linear momentum refers to the motion of an object in a straight line, while angular momentum refers to the motion of an object around an axis. Angular momentum depends on both the mass and the distribution of mass around the axis of rotation, while linear momentum only depends on the mass and velocity of an object.

5. How is linear motion related to force?

According to Newton's second law of motion, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This means that the greater the force applied to an object, the greater its acceleration and therefore its linear motion.

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