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What are the Practical Applications of Conservation of Linear Momentum?
Newton’s second law of motion states that, the rate of change of momentum of a body is directly proportional to the applied force. Hence, in the absence of an external force there is no change in the momentum of a body. This law is also applicable for a system consisting of a number of bodies. The members in a many-body system may have interactions among themselves due to collisions, attractions, repulsions etc. These forces are to be treated as internal forces, not external forces.
Statement: In the absence of any external force acting on a system of bodies, even if interactions exist among the bodies, the total linear momentum of the system remains constant.
This statement is the law of conservation of linear momentum. For a system of bodies, we can calculate the components of linear momentum, of all the bodies present, in any chosen direction. The sum of these individual components in this direction will be a constant.
We shall illustrate the law for a one-dimensional collision. Suppose two particles of mass m1 and m2, moving with velocities u1 and u2 respectively in a straight line, collide with each other [Fig.]. After collision, the particles move with velocities v1 and v2 respectively in the same direction.
Hence, total momentum before collision = m1u1 + m2, and total momentum after collision = m1v1 + m2v2.
If there is no external force, as per the conservation law of linear momentum,
m1u1 + m2u2 = m1v1 + m2v2 ……. (1)
Law of conservation of near momentum from Newton’s third law of motion:
i) In absence of any external force, two bodies, during a collision, exert an impulsive force on each other. From Newton’s third law of motion, impulsive force on the first body is equal and opposite to that exerted on the second body.
Forces F12 and F21 are shown in Fig.:
F21 = -F12.
If the collision lasts for a time t, the impulse of F21 on the first body = F21 ᐧ t = change of momentum of the first body = m1y1 – m1u1
Similarly, impulse of F12 on the second body = F12 × t = change of momentum of the second body = m2v2 – m2u2
As F21 = -F12, m1v1 – m1u1 = -(m2v2 – m2u2)
or, m1u1 + m2u2 = m1v1 + m2u2
As total momentum before collision = total momentum after collision, the total linear momentum remains conserved.
ii) Alternative method: Suppose two bodies of masses m1 and m2, moving along the same straight line with velocities u1 and u2 respectively, collide with each other.
Since there is no external force present, let the force on the first body exerted by the second body be F21, and that on the second body exerted by the first body be F12.
a1 = \(\frac{d v_1}{d t}\) and a2 = \(\frac{d v_2}{d t}\) are the respective accelerations of the bodies.
From Newton’s third law of motion,
F21 = – F12
or, m1\(\frac{d v_1}{d t}\) = -m2\(\frac{d v_2}{d t}\) or, \(\frac{d}{d t}\)(m1u1 + m2v2) = 0
or, m1v1 + m2u2 = constant
Hence, the total linear momentum remains conserved.
Newton’s third law of motion from the law of conservation of linear momentum: Let the initial momentum of two bodies be p1 and p2. They come in contact for a time t and their momenta change to p1’ and p2’ respectively.
In absence of any external force, as per law of conservation of linear momentum,
p1 + p2 = p1‘ + p2‘
or, p1‘ – p1 = -p2‘ + p2 or, \(\frac{p_1{ }^{\prime}-p_1}{t}\) = \(-\frac{p_2{ }^{\prime}-p_2}{t}\)
The left hand side of the equation is the rate of change of momentum of the first body = force on the first body -F21; similarly, the right hand side is -F12.
Hence F21 = -F12 or, action = – reaction.
So action and reaction between two bodies are equal and opposite. This is nothing but Newton’s third law of motion.
Practical applications of the principle of conservation of linear momentum:
i) Recoil of a gun: When a bullet is fired from a gun, the gun recoils or gives a kick in backward direction. It can be explained as follows:
let m1 be the mass of the bullet and m2 be the mass of the gun [Fig.]. Initially both the gun and the bullet are at rest. On firing the gun, let the bullet move with a velocity \(\vec{v}_1\) and the gun move with a velocity \(\vec{v}_2\). According to the prin-ciple of conservation of linear momentum, total momentum of gun and bullet before firing = total momentum of gun and bullet after firing.
i.e., 0 = m1\(\vec{v}_1\) + m2\(\vec{v}_2\) or, \(\vec{v}_2\) = \(-\frac{m_1}{m_2} \vec{v}_1\)
The negative sign shows that \(\vec{v}_2\) and \(\vec{v}_1\) are in opposite direction, i.e., as the bullet moves forward, the gun will move in backward direction. This backward motion of the gun is called recoil of the gun. Hence, while firing a bullet, the gun must be held tight to the shoulder—oth-erwise, because of the recoil velocity of the gun, the shoulder of the man who fires the gun, may get hurt. If the gun is held tight to the shoulder then the gun and the body of the man recoil as a single system. As the mass is quite large, the recoil velocity will be very small and the shoulder of the man will not get hurt.
ii) Explosion of a bomb: Let a bomb of mass M be initially at rest. So, the initial momentum = 0.
After the explosion, suppose that the bomb is split into a few fragments of masses m1, m2, m3, ……….; the fragments fly away from the centre of explosion with velocities \(\vec{v}_1\), \(\vec{v}_2\), \(\vec{v}_3\), ………. Thus, the final momentum is,
\(\vec{p}\) = \(\vec{p}_1\) + \(\vec{p}_2\) + \(\vec{p}_3\) + …… = m1\(\vec{v}_1\) + m2\(\vec{v}_2\) + m3\(\vec{v}_3\) + ……….
As no external force acts on the bomb, the momentum must be conserved, i.e.,
final momentum = initial momentum,
or, \(\vec{p}_1\) + \(\vec{p}_2\) + \(\vec{p}_3\) + ……. = 0
As a special case, if the bomb explodes into two fragments only, then
\(\vec{p}_1\) + \(\vec{p}_2\) = 0, or m1\(\vec{v}_1\) + m2\(\vec{v}_2\) = 0 , or m1\(\vec{v}_1\) = -m2\(\vec{v}_2\)
Again, if the two fragments are of equal mass,
m1 = m2 then \(\vec{v}_1\) = –\(\vec{v}_2\).
Hence, the two fragments would acquire equal and opposite velocities due to the explosion.
Numerical Examples
Example 1.
A bullet of mass 6 g is fired with a velocity of 500 m ᐧ s-1 from a gun of mass 4 kg. Find the recoil velocity of the gun.
Solution:
As the bullet and the gun were at rest before the firing, the initial momentum of the system was zero. Let the speed of the bullet after the firing be v, and that of the gun be V.
From the law of conservation of linear momentum, we get,
0 = MV + mv
[mass of the bullet = m, mass of the gun = M]
or, V = –\(\frac{m}{M} v\) = –\(\frac{6 \times 10^{-3}}{4}\) × 500 = -0.75 m ᐧ s-1
Hence, recoil velocity of the gun = 0.75 m ᐧ s-1.
Example 2.
While firing a bullet of mass 8 g, the recoil velocity of the gun of mass 5 kg becomes 64 cm ᐧ s-1. The bullet penetrates 50 cm through a target and then stops. Express the average resistance on the bullet in newton.
Solution:
Initial momenta of the gun and the bullet were zero, as both were at rest. Let the masses of the gun and the bullet be M and m respectively and the respective velocities after firing be V and v.
From the law of conservation of linear momentum,
o = MV + mv
∴ v = –\(-\frac{M}{m} V\) = \(-\frac{5}{8 \times 10^{-3}}\) × 0.64 = -400 m ᐧ s-1
The bullet comes to rest after penetrating a distance of 50 cm or 0.5 m. If the retardation is a due to the resistance of the material of the target,
0 = (400)2 – 2a ᐧ 0.5 or, a = 16 × 104 m ᐧ s-2
Hence average resistance,
F = ma = 0.008 × 16 × 104 = 1280 N.
Example 3.
A body of mass m moving with velocity V along the X-axis, collides with another mass M moving with velocity v along the Y-axis. The masses coalesce after collision. Find the velocity and the direction of motion of the combined mass.
Solution:
Let the velocity of the combined mass be u which makes an angle θ with the X-axis [Fig.].
Applying the law of conservation of linear momentum along the X-axis,
mV + 0 = (m + M)u cosθ
Similarly, along the Y-axis,
0 + Mv = (m + M)u sinθ
Squaring and adding (1) and (2),
Example 4.
A body of mass 50 kg is projected vertically upwards with a velocity of 100 m ᐧ s-1. After 5 sit splits up into two parts due to an explosion. One part of mass 20 kg moves vertically upwards with a velocity of 150 m ᐧ s-1. Find the velocity of the second body. Find the sum of momenta of the two parts 3 s after the explosion and show that if there was no explosion, the momentum of the body would have been constant. [HS 2000]
Solution:
Let the velocity of the projectile 5 s after the projection and just before the explosion be v.
∴ v = 100 – 9.8 × 5 = 51 m ᐧ s-1
Let the velocity of the second part after explosion be v1
Applying the law of conservation of linear momentum along the direction of projection,
50 × 51 = 20 × 150 + 30 × v1
or, v1 = \(\frac{50 \times 51-3000}{30}\) = -15 m ᐧ s-1 (downwards)
Let the velocity of the 20 kg mass, produced due to the explosion, 3 seconds after the explosion be v’.
∴ v’ = 150 – 9.8 × 3
= 150 – 29.4 = 120.6 m ᐧ s-1 (upwards)
If v” is the velocity of the 30 kg mass 3 seconds after the explosion, then
v” = 15 + 9.8 × 3
= 15 + 29.4 = 44.4 m ᐧ s-1 (downwards)
Thus the total momentum 3 seconds after the explosion
= 20 × 120.6 – 30 × 44.4 = 1080 kg.m ᐧ s-1
In the case of no explosion, the velocity after 8 seconds of projection would have been
v2 = 100 – 9.8 × 8 = 100 – 78.4 = 21.6 m ᐧ s-1
Hence, its momentum after 8 s would have been
50 × 21.6 = 1080 kg.m ᐧ s-1
Example 5.
A body P of mass 20 g and another body Q of mass 40g are projected at the same time from points A and B on the earth’s surface.
Velocity of projection for each was 49 m ᐧ s-1 and it was directed at an angle of 45° with the horizontal [Fig.]. Distance AB = 245m. P and Q colide on the same vertical plane. After collision, P retraces its path to the ground. Find the position where Q touches the ground. How long will Q take to reach the ground after the collision? [g = 9.8 m ᐧ s-1]
Solution:
As the two bodies P and Q are projected at the same time with the same velocity u = 49 ᐧ s-1 and angle = 45°, horizontal range for both P and Q is the same and equal to,
\(\frac{u^2 \sin 2 \alpha}{g}\) = \(\frac{(49)^2 \times \sin 90^{\circ}}{9.8}\) = 245 m.
Hence the two bodies P and Q will meet at a point on the perpendicular bisector of AB.
Let the time when P, Q meet after projection be t. Hence horizontal distance moved by each
x = 49 cos45° × t = \(\frac{245}{2}\), or, t = \(\frac{5}{\sqrt{2}} \mathrm{~s}\)
During the collision, vertical velocity of P and Q
uy = usin45° – gt = \(\frac{49}{\sqrt{2}}\) – \(\frac{9.8 \times 5}{\sqrt{2}}\) = 0
Horizontal velocity of P and Q
ux = ucos45° = \(\frac{49}{\sqrt{2}}\) m ᐧ s-1
Hence horizontal component of the total momentum before collision
Px = (0.02ux – 0.04ux)kg ᐧ m ᐧ s-1
P retraces its path after collision. Hence horizontal velocity of P after collision = -ux and for Q it is = v (say).
Hence horizontal component of the total momentum of the system after the collision
= (-0.02ux + 0.04v)kg ᐧ m ᐧ s-1
From the law of conservation of momentum,
0.02ux – 0.04 ux = (-0.02ux + 0.04v)
or, v = 0.
Thus Q does not have any horizontal component of velocity and hence it falls down vertically at the midpoint of AB, at a distance \(\frac{245}{2}\) = 122.5 m from both A and B
Let t = time required by Q to reach the ground after the collision.
t = \(\frac{u \sin 45^{\circ}}{g}\) = \(\frac{49 \times 1}{\sqrt{2} \times 9.8}\) = 3.53 s.
Example 6.
A car of mass 2000 kg collides with a truck of mass kg moving at 48 km ᐧ h-1. After collision the car rides up the truck and the truck-car combination moves at 15 km ᐧ h-1. What was the velocity of the car before collision?
Solution:
Suppose a body of mass m moving with velocity u collides along a straight line with a body of mass M and velocity v. After collision the two masses combine and move with velocity V. Applying the law of conservation of momentum,
mu + Mv = (m + M) V
or, mu = (m + M)V – Mv
or, u = \(\frac{m+M}{m} V\) – \(\frac{M}{m} \nu\) = (1 + \(\frac{M}{m}\))V – \(\frac{M}{m}\)v.
Here m = 2000 kg v = 48km ᐧ h-1, M = 10000 kg, V = 15km ᐧ h-1.
∴ \(\frac{M}{m}\) = \(\frac{10000}{2000}\)
u = (1 + 5) × 15 – 5 × 48 = 90 – 240 = -150 km ᐧ h-1
The negative sign indicates that before collision the car was moving in the direction opposite to that of the truck.
Example 7.
A ball weighing 100 g was thrown vertically upwards with a velocity 49 m ᐧ s-1. At the same moment another identical ball was dropped from a height of 98m vertically above the first ball. After some time the two balls collided and got stuck together. This combined mass reached the ground finally. Determine how long the balls were in motion.
Solution:
Let the height attained by the balls above the ground in time t1 be h when they collide with each other.
Analysing the upward motion of the first ball, we get
h = 49t1 – 9.8\(t_1^2 / 2\) ….. (1)
Analysing the downward motion of the second ball, we get
98 – h = 9.8\(t_1^2 / 2\) …… (2)
From equations (1) and (2)
98 = 49t1 or t1 = 2 s
At the time of collision the velocity of the ball thrown in the upward direction is v1 and that of the ball thrown in the downward direction is v2.
∴ v1 = 49 – 9.8 × 2 = 29.4 m (upward)
and v2 = 9.8 × 2 = 19.6 m ᐧ s-1 (downward)
After collision, the velocity of the combined mass is V. Then according to the law of conservation of momentum
0.1 × 29.4 – 0.1 × 19.6 = 2 × 0.1 × V
or, V = 4.9 m ᐧ s-1 (upward)
From equation (1), we get
h = 49 × 2 – 9.8 × (2)2/2 = 78.4 m
Let the time taken by the combined mass to reach the ground be t2. Then
78.4 = -4.9 t2 + 9.8 × \(t_2^2 / 2\) or, \(t_2^2\) – t2 – 16 = 0
or, t = (1±\(\sqrt{1+64}\))/2 = 4.53 s (∵ t ≮ 0)
∴ Total time = t1 + t2 = 2 + 4.53 = 6.53 s.
Example 8.
A body of mass m is at rest on a smooth horizontal plane. A force F = kt is applied on the body making an angle α with the horizontal. In the equation of force, t is time and k is a constant. At the instant the object loses contact with the plane, how far will it move along the plane and what will be its velocity?
Solution:
As F = kt, force increases with time. Vertical component of the force = Fsinα. The body loses contact with the plane when the vertical force, equals the weight of the body,
i.e. Fsinα = mg [Fig.]. Let the time after which the body loses contact be t0.
∴ Fsinα = kt0sinα = mg
∴ t0 = \(\frac{m g}{k \sin \alpha}\)
Horizontal component of the force F = Fcosα = ktcosα
Hence horizontal acceleration = \(\frac{k t \cos \alpha}{m}\) = \(\frac{d v_x}{d t}\)
[where B = integration constant]
As at the initial moment, the displacement is zero, i.e., at t = o, s = 0.
From equation (3), we get B = 0.
Inserting the value of B in equation (3),
Example 9.
A cannonball of mass 50 kg is fired with a velocity of 40 m ᐧ s-1 from a cannon of mass 1000 kg. What will be the recoil velocity of the cannon?
If the force of friction between the surface and the wheels of the cannon is \(\frac{1}{10}\)th of the weight of the cannon, how far will the cannon move before coming to rest? [Given, g = 10 m ᐧ s-2] [HS(old syllabus) ‘07]
Solution:
Let the velocity of recoil of the cannon be V.
Hence, from the law of conservation of momentum,
0 = mv + MV or, V = –\(\frac{m v}{M}\)
Here, m = mass of the cannonball = 50 kg, v = its velocity = 40 m ᐧ s-1 and M = mass of the cannon = 1000 kg.
From given data,
V = \(-\frac{50 \times 40}{1000}\) = -2 m ᐧ s-1
Hence, recoil velocity of the cannon = 2 m ᐧ s-1 (backward).
Frictional force between the wheels and the land surface
= \(\frac{1}{10}\) × 1000 × 10 = 1000N
∴ Retardation of the cannon = \(\frac{1000 \mathrm{~N}}{1000 \mathrm{~kg}}\) = 1 m ᐧ s-1
∴ Distance travelled by the cannon before coming to rest is,
s = \(\frac{V^2}{2 a}\) = \(\frac{2^2}{2 \times 1}\) = 2 m
Example 10.
Four identical blocks, each of mass m, are connected as shown in Fig. and are kept on a horizontal table. A force F is applied on the first block. Find the tension in each string, neglecting friction.
Solution:
Let the acceleration of the system, on applying force F on the first block, be a. Let tension in the string connecting the 1st and the 2nd block be T1, 2nd and that for the 3rd be T2 and for the 3rd and the 4th be T3 [Fig.].
Equations of motion for the
1st block, F – T1 = ma …. (1)
2nd block, T1 – T2 = ma …. (2)
3rd block, T2 – T3 = ma ….. (3)
and 4th block, T3 = ma ……… (4)
Since T3 = ma, T2 = ma + T3 = 2ma and T1 = 3ma.
Hence, applied force, F = ma + T1 = 4ma
Expressing the tensions in terms of the applied force F, we get
T1 = \(\frac{3}{4}\)F, T2 = \(\frac{F}{2}\) and T3 = \(\frac{F}{4}\)