Momentum Word Problems

J = Ft = mΔv =Δp **

 

Ex 1)  A 5.0 kg mass has its velocity change from 8.0 m/s east to 2.0 m/s east.  Find the objects change in momentum.

 

Ex 1)  A 5.0 kg mass has its velocity change from 8.0 m/s east to 2.0 m/s east.  Find the objects change in momentum.

m = 5.0 kg
Vi = 8.0 m/s East
Vf = 2.0 m/s East
Δp = ?

**J = Ft = mΔv =Δp **

 

Δp = mΔV

 

= (5.0 kg)(2.0 m/s - 8.0 m/s)

 

= -30. kgm/s East

 

or +30. kg m/s West


Ex 2) A 5.0 kg mass moving with a velocity of 8.0 m/s east has an impulse applied to it which causes its velocity to change to 20. m/s East.

 

Find Impulse:

 

Ex 2) A 5.0 kg mass moving with a vector of 8.0 m/s east has an impulse applied to it which causes its velocity to change to 20. m/s East.

Find Impulse:

m = 5.0 kg
Vi = 8.0 m/s East
Vf = 20. m/s East
J = ?

J = Ft = mΔv =Δp **


J = mΔv  = (5.0 kg)(12. m/s East)

= 60. kg m/s east = 60. Ns East

 

Find the force if the impulse was applied for 3.0 sec.

m = 5.0 kg
Vi = 8.0 m/s East
Vf = 20. m/s East
J = 60. kg m/s east

 

J = Ft = mΔv =Δp **

 

J = Ft = 60. Ns

 

F(3.0 sec) = 60. Ns

 

F = 20 N East

 

Ex 3) How long would it take for a net upward force of 100. N, to increase the speed of a 50. kg object from 100. m/s to 150. m/s.

 

Ex 3) How long would it take for a net upward force of 100. N, to increase the speed of a 50. kg object from 100. m/s to 150. m/s.

 

F = 100. N

m = 50. kg

Vi = 100. m/s East

Vf = 150. m/s East

t = ?

J = Ft = mΔv =Δp **

 

Ft  = mΔv

 

(100. N)t  =  50. kg(50. m/s)

 

t  =  25. seconds

 

Ex 4)  A 1.0 kg ball traveling @ 4.0 m/s strikes a wall and bounces straight back @ 2.0 m/s. 

ballRebound
 Racquetball Slow-Mo - HiViz

Find Δp 

Ex 4)  A 1.0 kg ball traveling @ 4.0  m/s strikes a wall and bounces straight back @ 2.0 m/s

Find Δp 

 

m = 1.0 kg

Vi = 4.0 m/s East

Vf = ?????

Vf = -2.0 m/s (opposite direction)

Δp = ?

J = Ft = mΔv =Δp **

(a) Δp = mΔv

= (1.0 kg)(-2.0 m/s - 4.0 m/s)

= - 6.0 kgm/s

 

(b) What is impulse applied to the ball?   

 

J =  Δp = - 6.0 kgm/s

 

(c) What is impulse applied to the wall?

 

J =  + 6.0 kgm/s

 

3rd Law, Action Reaction