With what speed does the cart recoil?

[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

vf^2=vi^2+2ad

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!

 

[Antarres]

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.

 

[Orodruin]

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?

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.

 

[Antarres]

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.

 

[posto002]

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.)

 

[Orodruin]

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.

 

[hutchphd]

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...

 

[TSny]

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

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

 

[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.

 

[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...

Yep, you're right. Thank you.