Calculating the velocity of a target and embedded projectile

In summary: Your calculation for the projectile acceleration should be (400-25)/0.1, not (25-400)/0.1. The final result is correct, just the calculation is wrong.
  • #1
NewtonIsAmazing
9
0

Homework Statement


A projectile of mass 100g, moving at a speed of 400m/s, collides with a stationary wooden target of mass 1.5kg which is free to move. The projectile enters the wooden target on impact and remains embedded in it.

Homework Equations



a) calculate the velocity of the target and embedded projectile immediately after the impact

b) calculate the average force of the impact, if the duration of the collision was 0.10 seconds

c) calculate the respective accelerations of the projectile and the target during the impact

The Attempt at a Solution



Attempt at a)

(0.1kg x 400m/s) + (1.5kg x 0m/s) = (0.1kg + 1.5kg) x V

40 kg m/s = 1.6kg x V

V = (40 kg m/s) / 1.6 kg

V = 25 m/s

Attempt at b) (I'm a little unsure of this question)

So I used this equation: f=(v-u)/t

F = ( 25 m/s - 400 m/s)/ 0.10 seconds

F = -3750N

Attempt at c) (the answer to this question is dependent on on b) being correct)

F=ma

a=F/m

acceleration of projectile =3750N / 0.1kg
acceleration of projectile = 37500m/s^2

Acceleration of target= 3750N/1.5kg
Acceleration of target= 2500m/s^2

I just know I've done something wrong. Can someone please help me?
 
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  • #2
NewtonIsAmazing said:

The Attempt at a Solution



Attempt at a)

(0.1kg x 400m/s) + (1.5kg x 0m/s) = (0.1kg + 1.5kg) x V

40 kg m/s = 1.6kg x V

V = (40 kg m/s) / 1.6 kg

V = 25 m/s
This looks fine. It is usually better to keep your equations in symbolic form while doing the algebra and only substitute numbers in at the end. It does not matter much when the equations are simple, but for more complex equations, you can more readily see simplifications. It's also usually less writing.
Attempt at b) (I'm a little unsure of this question)

So I used this equation: f=(v-u)/t
Where does the equation "f=(v-u)/t" come from? What are v and u? What are their units?
F = ( 25 m/s - 400 m/s)/ 0.10 seconds
If you divide meters per second by seconds, what would the units for the result be?
 
  • #3
jbriggs444 said:
This looks fine. It is usually better to keep your equations in symbolic form while doing the algebra and only substitute numbers in at the end. It does not matter much when the equations are simple, but for more complex equations, you can more readily see simplifications. It's also usually less writing.
Where does the equation "f=(v-u)/t" come from? What are v and u? What are their units?

I want to find the average force. V is the final velocity and U is the initial velocity. Their units are m/s

If you divide meters per second by seconds, what would the units for the result be?

Question c) is dependent on question b) being correct.

Is question b) correct? if not, what have I done wrong?
 
  • #4
NewtonIsAmazing said:
Attempt at a)

(0.1kg x 400m/s) + (1.5kg x 0m/s) = (0.1kg + 1.5kg) x V

40 kg m/s = 1.6kg x V

V = (40 kg m/s) / 1.6 kg

V = 25 m/s

This is correct.

NewtonIsAmazing said:
Attempt at b) (I'm a little unsure of this question)

So I used this equation: f=(v-u)/t

F = ( 25 m/s - 400 m/s)/ 0.10 seconds

F = -3750N

Have another look at the units of your calculation of the force - you missed something.

NewtonIsAmazing said:
Attempt at c) (the answer to this question is dependent on on b) being correct)

F=ma

a=F/m

acceleration of projectile =3750N / 0.1kg
acceleration of projectile = 37500m/s^2

Acceleration of target= 3750N/1.5kg
Acceleration of target= 2500m/s^2

I just know I've done something wrong. Can someone please help me?

The calculations are basically correct, but the values you gained in b) are wrong. Plus: There's an easier way to calculate the acceleration without using the results from b). What's the definition of acceleration in kinematics (not dynamics)?
 
  • #5
stockzahn said:
Have another look at the units of your calculation of the force - you missed something.

I'm not sure what I've missed
The calculations are basically correct, but the values you gained in b) are wrong. Plus: There's an easier way to calculate the acceleration without using the results from b). What's the definition of acceleration in kinematics (not dynamics)?

V = U + AT ?
 
  • #6
NewtonIsAmazing said:
I'm not sure what I've missed

Write down the units of your equation without the actual values.

NewtonIsAmazing said:
V = U + AT ?

Try to solve for A, maybe you'll see the answer then.
 
  • #7
stockzahn said:
Write down the units of your equation without the actual values.

N = (m/s - m/s)/s

Those are the units
Try to solve for A, maybe you'll see the answer then.

a = (v - u)/t

a = (25 - 400)/0.1

a = - 2750 m/s^2 (for the projectile)

a = (25 - 0) / 0.1
a = 250m/s^2 (for the target)

is this correct?
 
  • #8
NewtonIsAmazing said:
N = (m/s - m/s)/s

Those are the units

Does that seem correct to you? N = m/s2

NewtonIsAmazing said:
a = (v - u)/t

a = (25 - 400)/0.1

a = - 2750 m/s^2 (for the projectile)

a = (25 - 0) / 0.1
a = 250m/s^2 (for the target)

is this correct?

Except for a small mistake in the calculus for the projectile, yes.
 
  • #9
stockzahn said:
Does that seem correct to you? N = m/s2

So which values should I plug into N = m/s^2?
Except for a small mistake in the calculus for the projectile, yes.

What mistake did I make?
 
  • #10
NewtonIsAmazing said:
So which values should I plug into N = m/s^2?

What's the defintion of 1 N (force)? 1 N = 1 ...

NewtonIsAmazing said:
What mistake did I make?

(25 - 400) = ...
 
  • #11
stockzahn said:
What's the defintion of 1 N (force)? 1 N = 1 ...

I'm not sure
(25 - 400) = ...

So should it be 400-25 instead?
 
  • #12
NewtonIsAmazing said:
I'm not sure

I'm sure you find that within 30 seconds on the internet, if you don't know it.

NewtonIsAmazing said:
So should it be 400-25 instead?

No, the formula was correct, just the result was wrong.
 
  • #13
stockzahn said:
I'm sure you find that within 30 seconds on the internet, if you don't know it.

So 1N = 1 kg metres per second squared?
No, the formula was correct, just the result was wrong.

So what exactly did I get wrong in the working out?
 
  • #14
NewtonIsAmazing said:
So 1N = 1 kg metres per second squared?

Correct. Now try to use it for calculating b).

NewtonIsAmazing said:
So what exactly did I get wrong in the working out?

(25-400) = -375, not -275
 
  • #15
stockzahn said:
Correct. Now try to use it for calculating b).

Which kilogram should I use? The kg for the projectile or the target?
 
  • #16
NewtonIsAmazing said:
Which kilogram should I use? The kg for the projectile or the target?
Which mass you should use depends on which acceleration you calculated. Which acceleration did you calculate?
 
  • #17
In your initial attempt you wanted to calculat the force with difference in movement of the projectile... If you continue with that you have to use the projectile's mass. But according to Newton's 3rd law you get the same result calculating the force with all data of the wood.
 
  • #18
stockzahn said:
In your initial attempt you wanted to calculat the force with difference in movement of the projectile... If you continue with that you have to use the projectile's mass. But according to Newton's 3rd law you get the same result calculating the force with all data of the wood.

F=0.1kg(25m/s-400m/s)/0.1s
F=-375N

Is this the correct answer for b)?
 
  • #19
NewtonIsAmazing said:
F=0.1kg(25m/s-400m/s)/0.1s
F=-375N

Is this the correct answer for b)?

Yes, that's it.
 

1. How do you calculate the velocity of a target and embedded projectile?

The velocity of a target and embedded projectile can be calculated by using the formula v = d/t, where v is velocity, d is distance, and t is time. First, measure the distance traveled by the target and projectile. Then, record the time it took for the projectile to reach the target. Finally, divide the distance by the time to get the velocity.

2. What is the importance of calculating the velocity of a target and embedded projectile?

Calculating the velocity of a target and embedded projectile is important for understanding the speed at which the projectile is traveling and the impact it will have on the target. This information is crucial for accuracy and safety in various fields such as ballistics, sports, and transportation.

3. What are the units of measurement for velocity?

The units of measurement for velocity are typically meters per second (m/s), but it can also be measured in other units such as kilometers per hour (km/h) or feet per second (ft/s).

4. How does the mass of the projectile affect its velocity?

The mass of the projectile does not directly affect its velocity. However, a heavier projectile may require more force to reach the same velocity as a lighter projectile due to differences in inertia and air resistance. This is known as the principle of conservation of momentum.

5. What are some factors that can affect the accuracy of calculated velocity?

Some factors that can affect the accuracy of calculated velocity include human error in measurement, external forces such as wind and air resistance, and the precision of the equipment used to measure distance and time. It is important to take these factors into consideration when calculating velocity for more accurate results.

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