Download CHAPTER 13 LINEAR MOMENTUM AND IMPULSE

CHAPTER 14 LINEAR MOMENTUM AND IMPULSE
EXERCISE 79, Page 182
1. Determine the momentum in a mass of 50 kg having a velocity of 5 m/s.
Momentum = mass × velocity = 50 kg × 5 m/s = 250 kg m/s downwards
2. A milling machine and its component have a combined mass of 400 kg. Determine the
momentum of the table and component when the feed rate is 360 mm/min.
Momentum = mass × velocity = 400 kg ×
360 ×10−3
m/s = 2.4 kg m/s downwards
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3. The momentum of a body is 160 kg m/s when the velocity is 2.5 m/s. Determine the mass of the
body.
Momentum = mass × velocity
Hence,
from which,
160 = mass × 2.5
mass =
160
= 64 kg
2.5
4. Calculate the momentum of a car of mass 750 kg moving at a constant velocity of 108 km/h.
Momentum = mass × velocity
Mass = 750 kg and velocity = 108 km/h =
108
m/s = 30 m/s.
3.6
Hence, momentum = 750 kg × 30 m/s = 22500 kg m/s
5. A football of mass 200 g has a momentum of 5 kg m/s. What is the velocity of the ball in km/h.
Momentum = mass × velocity
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5 = 0.2 × v
Hence,
from which,
velocity, v =
5
= 25 m/s
0.2
= 25 × 3.6 km/h = 90 km/h
6. A wagon of mass 8 t is moving at a speed of 5 m/s and collides with another wagon of mass 12 t,
which is stationary. After impact, the wagons are coupled together. Determine the common
velocity of the wagons after impact.
Mass m 1 = 8 t = 8000 kg, m 2 = 12000 kg and velocity u 1 = 5 m/s, u 2 = 0.
Total momentum before impact = m 1 u 1 + m 2 u 2
= (8000 × 5) + (12000 × 0) = 40000 kg m/s
Let the common velocity of the wagons after impact be v m/s
Since total momentum before impact = total momentum after impact:
40000 = m 1 v + m 2 v
= v(m 1 + m 2 ) = v(20000)
Hence
v=
40000
= 2 m/s
20000
i.e. the common velocity after impact is 2 m/s in the direction in which the 8 t wagon is
initially travelling.
7. A car of mass 800 kg was stationary when hit head-on by a lorry of mass 2000 kg travelling at
15 m/s. Assuming no brakes are applied and the car and lorry move as one, determine the speed
of the wreckage immediately after collision.
Mass m 1 = 800 kg, m 2 = 2000 kg and velocity u 1 = 0, u 2 = 15 m/s
Total momentum before impact = m 1 u 1 + m 2 u 2
= (800 × 0) + (2000 × 15) = 30000 kg m/s
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Let the common velocity of the wagons after impact be v m/s
Since total momentum before impact = total momentum after impact:
30000 = m 1 v + m 2 v
= v(m 1 + m 2 ) = v(2800)
Hence
v=
30000
= 10.71 m/s
2800
i.e. the speed of the wreckage immediately after collision is 10.71 m/s in the direction in which
the lorry is initially travelling.
8. A body has a mass of 25 g and is moving with a velocity of 30 m/s. It collides with a second
body which has a mass of 15 g and which is moving with a velocity of 20 m/s. Assuming that
the bodies both have the same speed after impact, determine their common velocity (a) when the
speeds have the same line of action and the same sense, and (b) when the speeds have the same
line of action but are opposite in sense.
Mass m 1 = 25 g = 0.025 kg, m 2 = 15 g = 0.015 kg, velocity u 1 = 30 m/s and u 2 = 20 m/s.
(a) When the velocities have the same line of action and the same sense, both u 1 and u 2 are
considered as positive values
Total momentum before impact = m 1 u 1 + m 2 u 2 = (0.025 × 30) + (0.015 × 20)
= 0.75 + 0.30 = 1.05 kg m/s
Let the common velocity after impact be v m/s
Total momentum before impact = total momentum after impact
i.e.
1.05 = m 1 v + m 2 v = v(m 1 + m 2 )
1.05 = v(0.025 + 0.015)
from which, common velocity, v =
1.05
= 26.25 m/s in the direction in which the bodies
0.040
are initially travelling
(b) When the velocities have the same line of action but are opposite in sense, one is considered as
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positive and the other negative. Taking the direction of mass m 1 as positive gives:
velocity u 1 = +30 m/s and u 2 = - 20 m/s
Total momentum before impact = m 1 u 1 + m 2 u 2 = (0.025 × 30) + (0.015 × - 20)
= 0.75 - 0.30 = + 0.45 kg m/s
and since it is positive this indicates a momentum in the same direction as that of mass m 1 .
If the common velocity after impact is v m/s then
0.45 = v(m 1 + m 2 ) = v(0.040)
from which, common velocity, v =
0.45
= 11.25 m/s in the direction that the 25 g mass is
0.040
initially travelling.
9. A ball of mass 40 g is moving with a velocity of 5 m/s when it strikes a stationary ball of mass
30 g. The velocity of the 40 g ball after impact is 4 m/s in the same direction as before impact.
Determine the velocity of the 30 g ball after impact.
Mass m 1 = 40 g = 0.040 kg, m 2 = 30 g = 0.030 kg. Initial velocity u 1 = 5 m/s, u 2 = 0; final velocity
v 1 = 4 m/s, v 2 is unknown.
Total momentum before impact = m 1 u 1 + m 2 u 2
= (0.040 × 5) + (0.030 × 0) = 0.20 kg m/s
Total momentum after impact = m 1 v 1 + m 2 v 2
= (0.040 × 4) + (0.030 v 2 ) = 0.16 + 0.030 v 2
Total momentum before impact = total momentum after impact, hence
0.20 = 0.16 + 0.030v 2
from which, velocity of 30 g ball after impact, v 2 =
0.20 − 0.16
= 1.33 m/s
0.030
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10. Three masses, X, Y and Z, lie in a straight line. X has a mass of 15 kg and is moving towards Y
at 20 m/s. Y has a mass of 10 kg and a velocity of 5 m/s and is moving towards Z. Mass Z is
stationary. X collides with Y, and X and Y then collide with Z. Determine the mass of Z
assuming all three masses have a common velocity of 4 m/s after the collision of X and Y with
Z.
Mass m X = 15 kg, m Y = 10 kg, velocity u X = 20 m/s and u Y = 5 m/s.
Total momentum before X collides with Y = m X u X + m Y u Y
= (15 × 20) + (10 × 5) = 350 kg m/s
Let X and Y have a common velocity of v 1 m/s after impact.
Total momentum after X and Y collide = m X v 1 + m Y v 1
= v 1 (m X + m Y ) = 25v 1
Total momentum before impact = total momentum after impact, i.e. 350 = 25v 1 , from which,
common velocity of X and Y, v 1 =
350
= 14 m/s.
25
Total momentum after X and Y collide with Z = (m X + Y × 4) + (m Z × 4) (since the common velocity
after impact = 4 m/s)
= (25 × 4) + (4 m Z )
Total momentum before X and Y collide with Z = total momentum after X and Y collide with Z,
i.e.
(m X + Y × 14) = (25 × 4) + 4 m Z
i.e.
25 × 14 = 100 + 4 m Z
350 - 100 = 4 m Z
from which,
mass of Z, m Z =
250
= 62.5 kg
4
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EXERCISE 80, Page 185
1. The sliding member of a machine tool has a mass of 200 kg. Determine the change in
momentum when the sliding speed is increased from 10 mm/s to 50 mm/s.
Change of linear momentum = mass × change of velocity
change in momentum = 200 kg × (50 – 10) ×10 −3 m/s
Hence,
= 8 kg m/s
2. A force of 48 N acts on a body of mass 8 kg for 0.25 s. Determine the change in velocity.
Impulse = applied force × time = change in linear momentum
48 N × 0.25 s = mass × change in velocity
i.e.
= 8 kg × change in velocity
from which, change in velocity =
48 N × 0.25s
= 1.5 m/s (since 1 N = 1 kg m/s 2 )
8 kg
3. The speed of a car of mass 800 kg is increased from 54 km/h to 63 km/h in 2 s. Determine the
average force in the direction of motion necessary to produce the change in speed.
Change of momentum = applied force × time
i.e.
mass × change of velocity = applied force × time
i.e.
 63 54 
800 kg × 
−
 m/s = applied force × 2 s
 3.6 3.6 
800 ×
from which,
applied force =
2
9
3.6 = 1000 N or 1kN
4. A 10 kg mass is dropped vertically on to a fixed horizontal plane and has an impact velocity of
15 m/s. The mass rebounds with a velocity of 5 m/s. If the contact time of mass and plane is
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0.025 s, calculate (a) the impulse, and (b) the average value of the impulsive force on the plane.
(a) Impulse = change in momentum = m(u 1 - v 1 ) where u 1 = impact velocity = 15 m/s and
v 1 = rebound velocity = - 5 m/s (v 1 is negative since it acts in the opposite direction to u 1 )
Thus, impulse = m(u 1 - v 1 ) = 10 kg (15 - - 5) m/s = 10 × 20 = 200 kg m/s
(b) Impulsive force =
impulse 200 kg m / s
= 8000 N or 8 kN
=
time
0.025s
5. The hammer of a pile driver of mass 1.2 t falls 1.4 m on to a pile. The blow takes place in 20 ms
and the hammer does not rebound. Determine the average applied force exerted on the pile by
the hammer.
Initial velocity, u = 0, acceleration due to gravity, g = 9.81 m/s 2 and distance, s = 1.4 m.
Using the equation of motion: v 2 = u 2 + 2gs
v 2 = 0 2 + 2(9.81)(1.4)
gives:
from which,
impact velocity, v =
( 2)(9.81)(1.4) = 5.241 m/s
Neglecting the small distance moved by the pile and hammer after impact,
momentum lost by hammer = the change of momentum
= mv = 1200 kg × 5.241 m/s
Rate of change of momentum =
change of momentum 1200 × 5.241
=
= 314460 N
20 ×10−3
change of time
Since the impulsive force is the rate of change of momentum, the average force exerted on the
pile is 314.5 kN
6. A tennis ball of mass 60 g is struck from rest with a racket. The contact time of ball on racket is
10 ms and the ball leaves the racket with a velocity of 25 m/s. Calculate (a) the impulse, and
(b) the average force exerted by a racket on the ball.
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(a) Impulse = change of momentum = mv
= (0.060 kg)(25 m/s) = 1.5 kg m/s
(b) Impulsive force =
impulse 1.5 kg m / s
= 150 N
=
time
10 ×10−3
7. In a press-tool operation, the tool is in contact with the work piece for 40 ms. If the average force
exerted on the work piece is 90 kN, determine the change in momentum.
Change in momentum = applied force × time
= 90000 N × 40 × 10−3 = 3600 kg m/s
8. A gun of mass 1.2 t fires a shell of mass 12 kg with a velocity of 400 m/s. Determine (a) the
initial velocity of recoil, and (b) the uniform force necessary to stop the recoil of the gun in
150 mm.
Mass of gun, m g = 1.2 t = 1200 kg, mass of shell, m s = 12 kg,
and initial velocity of shell, u s = 400 m/s.
(a) Momentum of shell = m s u s = 12 × 400 = 4800 kg m/s.
Momentum of gun = m g v = 1200v where v = initial velocity of recoil of the gun.
By the principle of conservation of momentum, initial momentum = final momentum,
i.e.
0 = 4800 + 1200v
from which, velocity v =
− 4800
= - 4 m/s (the negative sign indicating recoil velocity)
1200
i.e. the initial velocity of recoil = 4 m/s
(b) The retardation of the recoil, a, may be determined using v 2 = u 2 + 2as, where v, the final
velocity, is zero, u, the initial velocity, is 4 m/s and s, the distance, is 150 mm, i.e. 0.15 m.
v 2 − u 2 02 − 42 −16
Rearranging v 2 = u 2 + 2as for a gives: a = = =
= - 53.33 m/s 2
2s
2 (0.15) 0.3
Force necessary to stop recoil in 150 mm = mass × acceleration
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= 1200 kg × 53.33 m/s 2
= 64000 N or 64 kN
9. In making a steel stamping, a mass of 100 kg falls on to the steel through a distance of 1.5 m and
is brought to rest after moving through a further distance of 15 mm. Determine the magnitude of
the resisting force, assuming a uniform resistive force is exerted by the steel.
v 2 = u 2 + 2gs, where v = final velocity, u = initial velocity = 0, g = 9.81 m/s 2
and s = distance = 1.5 m
Hence,
v 2 = 0 2 + 2(9.81)(1.5), from which,
velocity of mass, just before impact, v = 2(9.81)(1.5) = 5.425 m/s
Resistive force of ground = mass × acceleration.
The acceleration is determined using v 2 = u 2 + 2as where v = final velocity = 0,
u = initial velocity = 5.425 m/s and s = distance driven in ground = 15 mm = 0.015 m.
Hence,
0 2 = (5.425) 2 + 2(a)(0.015), from which,
acceleration, a =
− (5.425) 2
= - 981 m/s 2 (the minus sign indicates retardation)
2 (0.015)
Thus, resistive force of ground = mass × acceleration
= 100 kg × 981 m/s 2 = 98100 N or 98.1 kN
10. A vertical pile of mass 150 kg is driven 120 mm into the ground by the blow of a 1.1 t hammer
which falls through 800 mm. Assuming the hammer and pile remain in contact, determine
(a) the velocity of the hammer just before impact, (b) the velocity immediately after impact, and
(c) the resistive force of the ground, assuming it to be uniform.
(a) For the hammer, v 2 = u 2 + 2gs, where v = final velocity, u = initial velocity = 0, g = 9.81 m/s 2
and s = distance = 800 mm = 0.80 m
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Hence,
v 2 = 0 2 + 2(9.81)(0.80), from which,
velocity of hammer, just before impact, v = 2(9.81)(0.80) = 3.96 m/s
(b) Momentum of hammer just before impact = mass × velocity
= 1100 kg × 3.96 m/s = 4356 kg m/s
Momentum of hammer and pile after impact = momentum of hammer before impact.
Hence, 4356 kg m/s = (mass of hammer and pile) × (velocity immediately after impact)
i.e.
4356 = (1100 + 150)(v), from which,
velocity immediately after impact, v =
4356
= 3.48 m/s
1250
(c) Resistive force of ground = mass × acceleration. The acceleration is determined using
v 2 = u 2 + 2as where v = final velocity = 0, u = initial velocity = 3.48 m/s and s = distance driven
in ground = 120 mm = 0.12 m
Hence,
0 2 = (3.48) 2 + 2(a)(0.12), from which,
− (3.48) 2
acceleration, a =
= - 50.46 m/s 2 (the minus sign indicates retardation)
2 (0.12)
Thus, resistive force of ground = mass × acceleration
= 1250 kg × 50.46 m/s 2 = 63075 N = 63.08 kN
EXERCISE 81, Page 186
Answers found from within the text of the chapter, pages 180 to 185.
EXERCISE 82, Page 186
1. (d) 2. (b) 3. (f) 4. (c) 5. (a) 6. (c) 7. (a) 8. (g) 9. (f) 10. (f) 11. (b) 12. (e)
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