# With what speed does the cart recoil?

• posto002
In summary, the conversation discusses finding the initial velocity and speed of an object in projectile motion. The formula vf^2=vi^2+2ad is used, along with the conservation of momentum. However, there is some confusion about the correct mass to use in the calculations, with some suggesting it should be 0.54 kg instead of 0.59 kg. The final answer is determined to be v=0.71m/s.
posto002
Homework Statement
A system of inertia 0.59 kg consists of a spring gun attached to a cart and a projectile. The system is at rest on a horizontal low-friction track. A 0.050-kg projectile is loaded into the gun, then launched at an angle of 40∘ with respect to the horizontal plane. With what speed does the cart recoil if the projectile rises 2.1 m at its maximum height?
Relevant Equations

Conservation of momentum
I started off with vf^2=vi^2+2ad and plugged in the final velocity, acceleration (9.8) and distance which is 2.1 to get:
0=vi^2+2(9.8)(2.1) I solved this to get:
vi=6.42
I then plugged this into find the x-component of vi (I labeled this as simply x):
tan(40degrees)=6.42/(x) Solving for x I get:
x=7.65
After solving for the x-component, I use the conservation of momentum to produce:
0.59v=0.05(7.65)
0.59v=0.3825
v=0.65m/s for the speed that the object recoils, but that answer is incorrect. Can someone help me see what I did wrong? Thank you!

The law you used to find initial velocity, is law that works for accelerated motion in a straight line. Therefore, in case of projectile motion, it works on y component of the velocity, since acceleration is along that component. So by that formula, you get the y component of initial velocity, not the total initial velocity.

Your solution looks fine to me (apart from that gravitational acceleration should be -9.8 m/s^2 and you should use units everywhere). What is the suggested answer if you are told that is wrong?

Antarres said:
The law you used to find initial velocity, is law that works for accelerated motion in a straight line. Therefore, in case of projectile motion, it works on y component of the velocity, since acceleration is along that component. So by that formula, you get the y component of initial velocity, not the total initial velocity.
Which is why he used the tangent function and not the cosine when computing the horizontal component.

Orodruin said:
Which is why he used the tangent function and not the cosine when computing the horizontal component.
Oh damn, I'm blind...Excuse me then, in that case, the solution should be fine, except switching the sign for g.

Antarres said:
Oh damn, I'm blind...Excuse me then, in that case, the solution should be fine, except switching the sign for g.
I understand where switching the sign for g would be appropriate, but even with g being negative I still would get the same answer. Wouldn't I? I'll ask my professor about this and see whether it's the right answer or not. Thanks for your help (and for Orodruin's help.)

posto002 said:
but even with g being negative I still would get the same answer. Wouldn't I?
Your actual computed answer already assumed g to be negative. Otherwise you would have found an imaginary velocity component ...
In other words, you did the actual computation as if g was -9.8 m/s^2.

posto002
Did you count the projectile as part of the system mass? The problem is a bit ambiguous and the mass of the cart might be 0.54 kg...

posto002
Maybe you're supposed to interpret the 40o as the angle that the "gun" is tilted.

If so, then the initial velocity of the projectile will not be at 40o relative to the ground.
[EDIT: Later posts show that this is NOT how the problem statement is to be interpreted]

Last edited:
posto002
Okay, thank you all for your help. I reread the problem, and the entire process I went through was correct up until using the conservation of momentum when I used:
"0.59v=0.05(7.65)
0.59v=0.3825
v=0.65m/s "

Like hutchphd stated, the mass of the cart was 0.54 instead of 0.59 [taken from 0.59-0.050]. The final answer that I got from changing 0.59 to 0.54 was v=0.71m/s. Thank you once again for your help.

hutchphd and TSny
hutchphd said:
Did you count the projectile as part of the system mass? The problem is a bit ambiguous and the mass of the cart might be 0.54 kg...
Yep, you're right. Thank you.

hutchphd

## 1. What is meant by "cart recoil"?

"Cart recoil" refers to the backward motion of a cart after it has been pushed or pulled in a forward direction. It is a result of Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.

## 2. How is the speed of cart recoil calculated?

The speed of cart recoil can be calculated using the equation v = u + at, where v is the final speed, u is the initial speed, a is the acceleration, and t is the time. The acceleration in this case is equal to the force applied to the cart divided by its mass.

## 3. Does the weight of the cart affect its recoil speed?

Yes, the weight of the cart does affect its recoil speed. The heavier the cart, the greater the force needed to move it forward and the faster it will recoil. However, the mass of the cart will also affect the acceleration and time, which ultimately determine the final recoil speed.

## 4. Can the surface on which the cart is placed affect its recoil speed?

Yes, the surface on which the cart is placed can affect its recoil speed. A rough or uneven surface will create more friction, which can slow down the cart's recoil speed. On the other hand, a smooth and flat surface will have less friction and allow the cart to recoil faster.

## 5. How can the speed of cart recoil be measured?

The speed of cart recoil can be measured using a stopwatch and measuring the distance the cart travels in a given time. Alternatively, sensors and data collection systems can also be used to measure the speed of the cart with more accuracy.

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