### Impulse & Momentum Problems

1.  What is the impulse on a 2000 Kg mass if a force of 3257 N is applied to the mass in a forward direction?

1. a) Given: F = 3257 N                            Find:  Impulse (J)
m = 2000 Kg
t = 1.3 s

Solution:  J = Ft = 3257 N x 1.3 s = 4234 Ns = 4.2 x 103 Ns [Forward]

b) A pellet gun shoots a 30.0 g pellet at a speed of 200 m/s over a 50 ms interval.  Find the impulse.

b) Given : m = 30.0 g = 0.03 Kg        Find: Impulse (J)
v1 = 0 m/s
v2 = 200 m/s

t = 50 ms = 0.05 s

Solution:  J = p
Ft = mv
= 0.03 Kg (200 m/s)
= 6.0 Ns

c) A 500 g watermelon is dropped out of a window.  What impulse will the force of gravity acting on it for 3.0 s give the watermelon if free fall?

c) Given:  m = 500 g = 0.5 Kg                Find: Impulse (J)

t = 3.0 s
g = 9.8 m/s2

Solution:  J = p
Ft = J
= m x  g x
t
= 0.5 Kg x 9.8 N/Kg x 3.0 s
= 14.7 Ns [down] = 15 Ns [down]

2.  A 54 Kg blimp is traveling at 25 m/s [S] when it hits a tall water tower and bounces back at 20 m/s [N]. Find the blimp's change in momentum.

Given : m = 54 Kg        Find: (change in momentum)
v1 = 25 m/s [South]
v2 = 20 m/s [North] = -20 m/s [South]

Solution:  J = p
Ft = mv
= m (v2 -v1)
= 54 Kg (-20 - 25) m/s [S]
= -2430 Ns [S] = 2.4 x 103 Ns [N]

3. A 65 Kg mortar shell has an impulse of 2.5 x 103 N.s .

a) Find the force of the mortar on the shell if it takes 0.2 s to leave the barrel of the mortar.

Given:  J = 2.5 x 103 N.s               a) Find: F

t = 0.2 s                          b) Find: d
m = 65 Kg
v= 120 Km /h
=  33.3 m/s
v1 = 0 m/s

a) Solution:  J = p
Ft = J

\F = J / ∆t
= 2.5 x 103 Ns / 0.2 s
= 12 500 N = 1.3 x 104 N

b) What is the length of the barrel if the speed attained by the shell is 120 Km/h as it leaves the barrel?

b) First calculate the acceleration:  F = m x a
\a = F / m
= 1.25 x 104 N / 65 Kg = 192.3 m/s2

now calculate d using  v22 = v12  - 2ad
d = ( v22) / 2a = (33.3) / 385  m = 2.9 m

4. a) Given the following data for a Force-time graph. Plot the data and describe how to calculate the impulse.

 Force (N) 0 2 3 5 Time (s) 0 10 15 25

a) Solution: The impulse is given by the area under the Ft graph

\A = (25 x 5) / 2 = 52.5 Ns [S]

b)  how would you calculate the impulse if the graph is a non linear function of time such as y = F(t)

b) Solution: Find impulse by estimating the area under the F(t) graph using a series of small triangles or use calculus