Home

Momentum Word Problems

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
V_i = 8.0 m/s East
V_f = 2.0 m/s East
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 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
V_i = 8.0 m/s East
V_f = 20. m/s East
J = ?

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
V_i = 8.0 m/s East
V_f = 20. m/s East
J = 60. kg m/s east

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.

F = 100. N

m = 50. kg

V_i = 100. m/s East

V_f = 150. m/s East

t = ?

Ft  = mΔv

(100. N)t  =  50. kg(50. m/s)

t  =  25. secs

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

 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

V_i = 4.0 m/s E

V_f = ?

V_f = -2.0 m/s

(opposite direction)

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

Computer animation of a Newtons' cradle

DemonDeLuxe (Dominique Toussaint)

Law Of Conservation

 Of Momentum

© Tony Mangiacapre.,-All Rights Reserved [Home] Established 1995 - Use any material on this site (w/ attribution)