ENPH 131 Assignment #9
Solutions
Problem 15.2
The 11Mg “jump jet”is capable of taking off vertically from the deck of a ship.
a) If its jets exert a constant vertical force of 150kN on the plane, determine its velocity in t = 8 s starting from rest. Neglect
the loss of fuel during the lift.
Equation 15-4:
m Ivy M1 + ÚÙ Fy â t = m Ivy M2
0 + 150 000 H8.0L - H11 000L H9.81L H8.0L = 11 000 vy
vy = 30.6 m  s
b) Determine how high it goes in t = 8 s, starting from rest.
vy = vy0 + ay t
30.6 = 0 + ay H8.0L
ay = 3.826 m ‘ s2
s = s0 + v0 t +
1
2
a t2
1
s = 0 + 0 + 2 H3.826L H8.0L2
s = 122 m
Problem 15.26
The motor M pulls on the cable with a force of F, which has a magnitude that varies as shown on the graph.
2
Assignment 9.nb
If the 20kg crate is originally resting on the floor such that the cable tension is zero at the instant the motor is turned on,
determine the speed of the crate when t = 8 s. (Hint: First determine the time needed to begin lifting the crate.)
From 0 £ t £ 5s, F =
250
5
t = 50 t (N). The crate will begin to lift when the force of the motor F overcomes the force due to
gravity:
Fg = F
20 H9.81L = 50 t
t = 3.924 s
Apply eqn 15-4:
m Ivy M1 + ÚÙ Fy â t = mIvy M2
5
8
8
0 + Ù 50 t â t + Ù 250 â t - Ù H20L H9.81L â t = 20 v
v=
3.924
1
A25
I52
20
5
3.924
- 3.9242 M + 250 H8 - 5L - H20L H9.81L H8 - 3.924LE
v = 9.52 m  s
Problem 15.42
The 66kg boy leaps off cart A with a horizontal velocity of v ' = 2 m  s measured relative to the cart. Carts A and B have the
same mass of 45kg and are originally at rest.
a) Determine the velocity of cart A just after the jump.
Relative velocity equation for the boy and cart A:
vm = vA + vmA
Hvm L2 = HvA L2 + 2
Assignment 9.nb
3
[1]
Conservation of linear momentum:
Úm1 v1 = Úm2 v2
0 = mm Hvm L2 + mA HvA L2
m
Hvm L2 = - m A HvA L2
m
[2]
Solving [1] and [2] yeilds
Hvm L2 = 0.8108 m  s
HvA L2 = -1.189 m  s = -1.19 m  s
b) If he then lands on cart B with the same velocity that he left cart A, determine the velocity of cart B just after he lands.
Conservation of linear momentum assuming cart B is initially at rest:
Úm1 v1 = Úm2 v2
mm vm = Hmm + mB L v
H66L H0.8108L = H66 + 45L v
v = 0.482 m  s
Problem 15.53
The 22lb cart B is supported on rollers of negligible size.
a) If a 6lb suitcase A is thrown horizontally onto the cart at 10ft/s when it is at rest, determine the length of time that A slides
relative to B. The coefficient of kinetic friction between A and B is Μk = 0.4.
Conservation of linear momentum of entire system:
Úm1 v1 = Úm2 v2
6+22
10 + 0 = I 32.2 M v
v = 2.143 ft  s
6
32.2
The system’s speed after the case stops sliding is 2.143m/s
For A:
m v1 + ÚÙ F â t = m v2
6
32.2
6
10 - H6L H0.4L t = I 32.2 M H2.143L
t = 0.610 s
4
Assignment 9.nb
b) Determine the final velocity of A and B.
From above:
v = 2.14 ft  s
Problem 15.59
The 2kg ball is thrown at the suspended 22kg block with a velocity of 4m/s. If the coefficient of restitution between the ball
and the block is 0.6, determine the maximum height h to which the block will swing before it momentarily stops.
Conservation of linear momentum:
Úm1 v1 = Úm2 v2
mA vA1 + mB vB1 = mA vA2 + mB vB2
H2L H4L + 0 = H2L HvA2 L + H22L HvB2 L
[1]
Coefficient of restitution:
vB2 -vA2
vA1 -vB1
v -v
0.6 = B24-0A2
e=
H0.6L H4L = vB2 - vA2
[2]
Solving [1] and [2]:
vA 2 = -1.8667 m  s
vB 2 = 0.533 m  s
Now use conservation of energy for the block:
1
2
T1 + V1 = T2 + V2
H22L H0.533L2 + 0 = 0 + H22L H9.81L h
h = 0.0145 m = 14.5 mm