# Homework Help: Momentum/Impulse Question

1. Oct 21, 2007

### Draco

1. The problem statement, all variables and given/known data
A firecracker in a coconut blows the coconut into three pieces. Two pieces of equal mass fly off south and west, perpendicular to each other, at 20m/s. The third piece has twice the mass as the other two.

2. Relevant equations
p=mv
pi=pf

3. The attempt at a solution
My attempt involved giving the variable m to the two equal pieces and 2m to the third piece. I gave a value to m. Also the momentum before and after is equal to zero so I worked out the momentum and re arranged the equation to solve for v. I got 20m/s but i was wrong.

1. The problem statement, all variables and given/known data
Far in space, where gravity is negligible, a 495 kg rocket traveling at 95.0 m/s fires its engines. The figure View Figure shows the thrust force as a function of time. The mass lost by the rocket during these 30 s is negligible.

http://img212.imageshack.us/img212/8301/knightfigure0926xe8.th.jpg [Broken]
1)What impulse does the engine impart to the rocket?
2)At what time does the rocket reach its maximum speed?
3)What is the maximum speed of the rocket?

2. Relevant equations
Impulse= (Force)(Change in Time)

3. The attempt at a solution
Um I tried finding the area under the graph to find the impulse but that didn't work. I'm really having a hard time trying to read the graph.

Any tips for a struggling student? :S

Last edited by a moderator: May 3, 2017
2. Oct 21, 2007

### Shooting Star

1. You don't have to ascribe a value to m, thta'll just make it more messy. The m will cancel out. Equate the x and y components of the momentum. Try again, you must have made some error.

2. Area under graph = impulse. Remember something else which impulse is? It's equal to change in...

Last edited: Oct 21, 2007
3. Oct 21, 2007

### Draco

1) I'm not sure exactly what you mean. Can you be a bit clearer?

2) isnt it change in momentum? but like I'm not sure what to do with that.

4. Oct 21, 2007

### Shooting Star

1. The total momentum is zero, initially and finally. If v is the velo of the bigger piece of mass 2m, then its momentum is 2mv. Choose +ve x-axis in the direction east, and +ve y-axis towards north.
Then, if â€˜bâ€™ is the angle of the direction in which the mass 2m is going with the +ve x-axis, its x component of the momentum is 2mv*(cos b). The x component of the other two pieces are: 0 and (â€“m*20). So, 2mv*(cos b) â€“20m+0=0. Similarly, for the y component. Then, you have two equations and you can solve for v. (Here, by symmetry, the bigger piece must be going in the NE direction, and so b=45 degree.)

2. If you know the mass of the rocket, and the change in momentum, then you should be able to find the change in velo. The area under the graph gives you the total impulse, which is the change in momentum.

When do you think the speed of the rocket is max? When is the acceleration max?

5. Oct 21, 2007

### Shooting Star

>"I'm really having a hard time trying to read the graph."

The graph is simple. Just find the area as you would do in a triangle. Take the numbers on the axes as given. So, you'll be multiplying some seconds by some newtons to get the value of the impulse.

6. Oct 21, 2007

### Draco

ok so i got the maximum speed, but im still not sure how to find the time. Shouldnt the rocket be at maximum speed when it reaches the highest force?

7. Oct 22, 2007

### Shooting Star

Tell me when it reaches the max speed.

Suppose you smack a ball and just let it move in space without friction and no other force is acting on it, do you think the ball slows down?

Also, tell me when is the accelerarion max.

8. Oct 23, 2007

### Shooting Star

In case you are still not very sure, the max speed is reached at the end of 30 s, when the engine has been switched off. As long as there is a thrust, the speed will continue to increase.

It is the accn which is max when the force is max, i.e., at 10 s.