A 2kg rock is moving at a speed of 6m/s. What constant force is needed to stop the rock in 7 x 10^-4?

Explanation:

key to this problem is the impulse-momentum theorem which states that the change in the momentum of an object is equal to the impulse applied into it.

J

=

Δ

p

,

where

J

is the impulse and

Δ

p

is the change in momentum. Basically, the impulse is the product of force and time duration, that is,

J

=

F

Δ

t

In this problem, the impulse would be the product of the force stopping the rock and

0.7

s

.

On the other hand, momentum

p

is the product of the mass

m

and velocity

v

. Therefore, the change in momentum is given by

Δ

p

=

m

2

v

2

−

m

1

v

1

.

Starting with the impulse-momentum equation, we have

J

=

Δ

p

F

Δ

t

=

m

2

v

2

−

m

1

v

1

Divide both sides by

Δ

t

,

we get

F

Δ

t

Δ

t

=

m

2

v

2

−

m

1

v

1

Δ

t

F

=

m

2

v

2

−

m

1

v

1

Δ

t

Finally, substitute the values and we get

F

=

(

2

kg

)

(

0

)

−

(

2

kg

)

(

6

m

s

)

(

0.7

s

)

F

≈

−

20

kg

m

s

2

Since

1

N

=

1

kg

m

s

2

,

then

F

≈

−

20

N

Therefore, using the correct significant figures (in this case, we need one significant figure since 2 kg, 6 m/s and 0.7 s all have one) in the final answer, we would need to have approximately

20

N

force to stop the rock in

0.7

s

.

Note: The negative sign is referring to the direction of the force opposite of the direction of the velocity

v

1

.

0.19 m/s and 0.45 m/s