Velocity after force applied for 0.01 s

[Anjum S Khan]

Homework Statement

A ball(1kg) at rest was hit by a stick to set it in motion. Assuming Force (10N) was applied by a stick, and stick remained in contact with the ball for 0.01s. Ball moves from A to B (10m) in time t.
Find,
a) Velocity at B ?
b) Time t ?
(assume frictionless surface)

Homework Equations

F = ma, V = U + aT, V² = U² + 2as

The Attempt at a Solution

a = F/m = 10N/1kg = 10ms^-2, how to proceed after this ?

 

[drvrm]

Homework Statement

A ball(1kg) at rest was hit by a stick to set it in motion. Assuming Force (10N) was applied by a stick, and stick remained in contact with the ball for 0.01s. Ball moves from A to B (10m) in time t.
Find,
a) Velocity at B ?
b) Time t ?
(assume frictionless surface)

Homework Equations

F = ma, V = U + aT, V² = U² + 2as

The Attempt at a Solution

a = F/m = 10N/1kg = 10ms^-2, how to proceed after this ?

i think for such short duration force acting the concept of impulsive force and its effect on the state of motion/rest must be considered.

 

[Chestermiller]

You know the acceleration and you know the amount of time that the force is applied. So, what is the velocity after acceleration stops?

 

[Anjum S Khan]

You know the acceleration and you know the amount of time that the force is applied. So, what is the velocity after acceleration stops?

Velocity will be V = 10*0.01 = 0.1ms^-1.
Now, this velocity will be come initial velocity for motion A to B.

And since surface is frictionless, so

V_b = U_a = 0.1ms^-1.

And as velocity will be constant from A to B, so t = AB/U_a = 10m/0.1ms^-1 = 100sec.

 

[UchihaClan13]

Let's visualise what's happening
The stick remains in contact with the ball for 0.01 seconds
So the impulse transferred equals force exerted times the duration=0.1kgm/s
Now before contact , as the ball was initially at rest
The velocity of the ball when it loses contact with the bat/stick
Equal to 0.1/1=0.1m/s
Therefore this becomes your initial velocity of travel
Now remember that the 10N force no longer acts and thus there is no acceleration and due to the surface being frictionless, the ball moves with a constant velocity of 0.1m/s
When the distance is divided by this, we get the time taken
Which equals 10/0.1=100 seconds!
Your approach is absolutely correct!
UchihaClan13

 

[Anjum S Khan]

Thanks @UchihaClan13 and @Chestermiller for your efforts.

 

[Chestermiller]

In Physics Forums, if you want to thank someone, all you need to do is check the "like" icon on the bottom of one of their posts. It's much less effort than writing a separate post.

 

[CWatters]

If the ball starts at A then the answer should probably state that the duration of the acceleration phase is << than the coasting phase so it's effect on the time is negligible and can be ignored. It's not always true.