What is Bob's recoil speed on frictionless ice?

[sona1177]

Bob, who has a mass of 75 kg, can throw a 500 g rock with a speed of 30 m/s. The distance through which his hand moves as he accelerates the rock forward from rest until releases it is 1 m.

A) what constant force must Bob exert on the rock to throw it with this speed?

B) if bob is standing on frictionless ice, what is his recoil speed.

I got part A by doing V^2= Vi^2 + 2a(Xf-Xi), where 30=Vf, 0=vi, Xf-Xi= 1. And I got 450 m/s^2 for the acceleration. Then I plugged this in F=ma=.50 * 450= 230 N

I have no idea how to do part B. The only force I see on bob are the normal force and weight and perhaps the force of his legs propelling him back? I'm not sure! Can someone please help guide me for part B?

Thanks!

 

[sir_manning]

Your procedure for part A is correct and you have the right acceleration, but check your force calculation again. As for part B, remember Newton's third law? That's the one that says that if Bob pushes on something with a force, that something will push right back on bob with the same force.

 

[Bartek]

I have no idea how to do part B. The only force I see on bob are the normal force and weight and perhaps the force of his legs propelling him back? I'm not sure! Can someone please help guide me for part B?

Thanks!

First, 0.5*450 is 225 not 230

When Bob throwing rock - the rock throwing Bob according 3th Newtons law. That force accelerating Bob.

Because of force is a vector, it is important in what direction Bob throwing rock. Assume horizontally.

You should use momentum conservation law. Consider initial and final state of Bob-rock system.

regards
Bartek
ps
WOW. 30m/s! If Bob throws that rock on 45 deg it will flying for over 90 meters! He is not Bob, he is Superbob!

 

[sona1177]

Sorry! The force calculation should be 225 N. Ok so for part b I did 225=75a and got a=3 and then plugged this into V^2=Vi^2 + 2aX and got 30 m/s where Vi was 30 m/s (velocity he started at after leaving the ball), a=3, X=1. So my answer is 30 m/s but the books says it's .20m/s. I'm very confused.

 

[sona1177]

sona1177 Says: 05:19 AM


Sorry! The force calculation should be 225 N. Ok so for part b I did 225=75a and got a=3 and then plugged this into V^2=Vi^2 + 2aX and got 30 m/s where Vi was 30 m/s (velocity he started at after leaving the ball), a=3, X=1. So my answer is 30 m/s but the books says it's .20m/s. I'm very confused.

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[sona1177]

I haven't learned momentum conservation law, my book doesn't cover it until chapter 10 and I'm only on chapter 5
Right now :(

 

[Bartek]

I haven't learned momentum conservation law, my book doesn't cover it until chapter 10 and I'm only on chapter 5
Right now :(

I'm, confused too...

OK.. let's try without momentum.

According 3th Newton's law Bob->rock force is equal to rock->Bob force, is it cleary?

So you can calculate separately Bob's and rock acceleration using 2nd Newton's law. When you use acceleration's definition (a=(Vf-Vi)/t) and divide Bob's acceleration by rock's acceleration you can calculate Bob's velocity.

(You can asssume that you know time... when you divide accelerations it will be canceled).

 

[sona1177]

Why would I have to divide their accelerations? I still don't understand! :(

 

[Bartek]

Why would I have to divide their accelerations? I still don't understand! :(

This is the only way to solve problem if we can't use momentum conservation. It is just hint.

We have only 3th Newtons law with forces equality and 2nd law with f=ma equation. We have also a=(Vf-Vi)/t equation. We have no time and we have no rock's velocity in Bob's reference system!

When you divide accelerations, time and force will be canceled out. So you will have velocities ratio. That ratio will depends only from masses. Try.

EDIT: OK, I agree, that calculate this ratio isn't clear. I think that this equation is more natural: F_bob=F_rock. Then M_Bob*a_Bob=M_rock*a_rock. When you use acceleration def. (a=(Vf-Vi)/t) you can solve this.

 

[dulrich]

Sorry! The force calculation should be 225 N. Ok so for part b I did 225=75a and got a=3...

This part is right. Bob's acceleration is 3 m/s² in the opposite direction.

...and then plugged this into V^2=Vi^2 + 2aX and got 30 m/s where Vi was 30 m/s (velocity he started at after leaving the ball), a=3, X=1. So my answer is 30 m/s but the books says it's .20m/s. I'm very confused.

But this assumes that x = 1 meter for Bob, which is not necessarily true. He moves his hand 1 meter to launch the rock, but his body does not move 1 meter -- which is what you need to put into this formula to get the right answer. So how far does he move? You don't know. You need a different approach.

The one thing that is the same for both the rock and Bob is that the force on each is applied for the same amount of time. So how long does Bob spend launching the rock? (Hint: you need to use a formula that involves velocities, acceleration, and time). Use that time again with Bob's acceleration to find his final velocity. You should get the 0.2 m/s the book is quoting.