Force required to change direction

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rasen58
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Homework Statement


The speed of a 2 kg object changes from 30 m/s to 40 m/s during a 5 s time interval. During this same time interval, the velocity of the object changes in direction by 90 degrees. What average force was applied during the time interval?

Homework Equations


impulse = F.t

The Attempt at a Solution


I know how to find the force if the object hadn't turned.
2 kg (40 - 30) = 20 = F . t
F = 4 N
But I don't know how to find the force required to move it.

The answer is apparently 50N.
 
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Velocity is vector quantity. Draw the initial and final velocities and their difference: They make a right triangle. You need the magnitude of v2-v1.
 

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So v2-v1 is 50 m/s. But then what do I do. Finding the impulse by using (2 kg)(50m/s - 30m/s) doesn't give me the right answer.
 
Take a look at the change in momentum that has to occur in the x and y directions separately. Suppose the object is initially traveling in the +x direction. What's its momentum? After the maneuver, what's the new momentum in the x direction? So the change is? Then do the same for the y-direction.

You'll then have two components of the total change in momentum: ##ΔP_x## and ##ΔP_y##. ##ΔP## is a vector quantity. Can you determine the average force required from there?
 
rasen58 said:
So v2-v1 is 50 m/s. But then what do I do. Finding the impulse by using (2 kg)(50m/s - 30m/s) doesn't give me the right answer.
50 m/s is the magnitude of the difference between the initial and final velocities. Why do you subtract the initial speed again?
Multiplied by the mass, you get the change of momentum. Divided by time, it yields the average force.
 
ehild said:
50 m/s is the magnitude of the difference between the initial and final velocities. Why do you subtract the initial speed again?
Multiplied by the mass, you get the change of momentum. Divided by time, it yields the average force.
Right, but it's not going to give 50N. rasen58, are you sure it's 5 sec, not 2 sec?