## help

tizihuxi
Posts: 6
Joined: Tue Jun 05, 2018 9:50 pm

### help

A fan cart with a mass m1 starts from rest and travels a distance d on a frictionless track in time t1. How much time, t2, will the fan cart take to travel the same distance (starting from rest) if an additional weight of mass m2 is added to the fan cart?

Is the answer t1=t2? I think what I am doing is not right. I know that somehow the kinematic equations need to come into play, but don't understand how. Both velocity and acceleration are equal to 0 at the start.

jeff
Posts: 38
Joined: Mon Mar 05, 2018 11:54 pm

### Re: help

Hi tizihuxi,
tizihuxi wrote:
Tue Jun 05, 2018 9:52 pm
A fan cart with a mass m1 starts from rest and travels a distance d on a frictionless track in time t1. How much time, t2, will the fan cart take to travel the same distance (starting from rest) if an additional weight of mass m2 is added to the fan cart?

Is the answer t1=t2? I think what I am doing is not right. I know that somehow the kinematic equations need to come into play, but don't understand how. Both velocity and acceleration are equal to 0 at the start.
No, in this case t1 will not be equal to t2. There actually is a video of this experimental setup here:

You can see from that video that the added mass will make the cart have a smaller accelerate, since the force is essentially constant.

So there are several ways we can approach this, but the main idea is that we need to relate the various quantities. In this case, we want to determine how time depends on mass, and it is the force and distance that are the same for each cart.

Can you see how to use Newton's Law and a kinematic equation to get an equation with force, distance, time, and mass? (when initial velocity is zero)

tizihuxi
Posts: 6
Joined: Tue Jun 05, 2018 9:50 pm

### Re: help

Thanks for the video. Yes you are right, the time will be different.

Would the equation be F=ma and d=1/2at^2?

jeff
Posts: 38
Joined: Mon Mar 05, 2018 11:54 pm

### Re: help

tizihuxi wrote:
Wed Jun 06, 2018 1:40 am
Thanks for the video. Yes you are right, the time will be different.

Would the equation be F=ma and d=1/2at^2?
That's right, those are the ones we need. Notice in those, that acceleration is in both equations, and we don't want it in our formula. (It's not the same for both systems, and we don't care about it here.)

If you solve one of those equations for acceleration a and plug it in the other, what do you get?

Once you have that, you can use it to relate t1 and t2 together.

tizihuxi
Posts: 6
Joined: Tue Jun 05, 2018 9:50 pm

### Re: help

Well, it's not too clear to me. But, I think it is

d=1/2(F/m)t^2

So, I need this equation for the two carts. The only thing I am not sure is how to solve it. How would you go about it?

jeff
Posts: 38
Joined: Mon Mar 05, 2018 11:54 pm

### Re: help

tizihuxi wrote:
Wed Jun 06, 2018 5:25 pm
Well, it's not too clear to me. But, I think it is

d=1/2(F/m)t^2

So, I need this equation for the two carts. The only thing I am not sure is how to solve it. How would you go about it?
Yes, that's right! and we can rewrite it so that everything that is the same for both is on one side, like this:

t^2/m = 2d/F

Now this is for any cart, right? So for the first cart, its time is t1 and its mass is m1, so:

t1^2/m1 = 2d/F

and for the second cart, its time is t2 and its mass is m1+m2, so:

t2^2/(m1+m2) = 2d/F

Now d and F is the same for both carts, so the right hand side of those two equations are equal to each other, which means the left hand sides must be equal to each other as well:

t1^2/m1 = t2^2/(m1+m2)

and we can solve that for t2! Does that make sense? what do you get for that?

tizihuxi
Posts: 6
Joined: Tue Jun 05, 2018 9:50 pm

### Re: help

Thank you for your help. I managed to finish the last part.