EXPLANATION OF CONSERVATION OF MOMENTUM

EXPLANATION OF CONSERVATION OF MOMENTUM

Suppose two objects (two balls A and B, say) of masses m_A and m_B are travelling in the same direction along a straight line at different velocities u_A and u_B, respectively [Fig. 9.14(a)].

And there are no other external unbalanced forces acting on them. Let u_A > u_B and the two balls collide with each other as shown in Fig. 9.14(b). During collision which lasts for a time t, the ball A exerts a force F on ball B and the ball B exerts a force F_BA on ball A. Suppose v_A and v_B are the velocities of the two balls A and B after the collision, respectively [Fig. 9.14(c)]

Figure 9.14: Conservation of momentum in the collision of two balls.

From Eq.(9.1), the momenta (plural of momentum) of ball A before and after the collision are m_Au_A and m_Av_A, respectively. The rate of change of its momentum (or F_AB) during the collision will be 

\(m_A \frac{(v_A-u_A)}{t}\)

Similarly, the rate of change of momentum of ball B (= F_BA) during the collision will be

\(m_B \frac{(v_B-u_B)}{t}\)

According to the third law of motion, the force F_AB exerted by ball A on ball B and the force F_BA exerted by the ball B on ball A must be equal and opposite to each other. Therefore,

F_AB = - F_BA                                            _______(9.6)

or _{\(m_A\frac{(v_A-u_A)}{t}=-\,m_B\frac{(v_B-u_B)}{t}\)}                            

This gives,

m_Au_A  + m_Bu_B  = m_Av_A + m_Bv_B     _______(9.7)

Since (m_Au_A + m_Bu_B ) is the total momentum of the two balls A and B before the collision and (m_Av_A + m_Bv_B) is their total momentum after the collision, from Eq. (9.7) we observe that the total momentum of the two balls remains unchanged or conserved provided no other external force acts.

As a result of this ideal collision experiment, we say that the sum of momenta of the two objects before the collision is equal to the sum of momenta after the collision provided there is no external unbalanced force acting on them. This is known as the law of conservation of momentum. This statement can alternatively be given as the total momentum of the two objects is unchanged or conserved by the collision.

Source: This topic is taken from NCERT TEXTBOOK

Activity on Third Law of Motion

Activity 9.5: ( Balloon Rocket )

*   Take a big rubber balloon and inflate it fully. Tie its neck using a thread. Also using adhesive tape, fix a straw on the surface of this balloon.

*   Pass a thread through the straw and hold one end of the thread in your hand or fix it on the wall.

*   Ask your friend to hold the other end of the thread or fix it on a wall at some distance. This arrangement is shown in Fig. 9.15.

*   Now remove the thread tied on the neck of the balloon. Let the air escape from the mouth of the balloon.

*   Observe the direction in which the straw moves.

Figure 9.15: 

Activity 9.6: ( Recoil of Test Tube )

*   Take a test tube of good quality glass material and put a small amount of water in it. Place a stop cork at the mouth of it.

*   Now suspend the test tube horizontally by two strings or wires as shown in Fig. 9.16.

*   Heat the test tube with a burner until water vaporises and the cork blows out.

*   Observe that the test tube recoils in the direction opposite to the direction of the cork.

Figure 9.16: 

*   Also, observe the difference in the velocity the cork appears to have and that of the recoiling test tube.

Illustration 9.6:

A bullet of mass 20 g is horizontally fired with a velocity of 150 ms^-1 from a pistol of mass 2 kg. What is the recoil velocity of the pistol?

Sol:

We have the mass of bullet, m₁ = 20 g (= 0.02 kg) and the mass of the pistol, m₂ = 2 kg; initial velocities of the bullet (u₁) and pistol (u₂) = 0, respectively.

The final velocity of the bullet, v₁ = + 150 ms^-1.

The direction of the bullet is taken from left to right (positive, by convention, Fig. 9.17).

Figure 9.17:  Recoil of a pistol

Let v be the recoil velocity of the pistol

Total momenta of the pistol and bullet before the fire, when the gun is at rest

= (2 + 0.02) kg\(\times\)0 ms^–1

= 0 kgms^–1

Total momenta of the pistol and bullet after it is fired

= 0.02 kg \(\times\) (+ 150 ms^–1) + 2 kg \(\times\) v ms^–1

= (3 + 2v) kgms^–1

According to the law of conservation of momentum

Total momenta after the fire = Total momenta before the fire

3 + 2v = 0

\(\Rightarrow\) v = - 1.5 ms^–1.

The negative sign indicates that the direction in which the pistol would recoil is opposite to that of a bullet, that is, right to left.

Activity 9.7 and 9.8

Illustration 9.7:

A girl of mass 40 kg jumps with a horizontal velocity of 5 ms^-1 onto a stationary cart with frictionless wheels. The mass of the cart is 3 kg. What is her velocity as the cart starts moving? Assume that there is no external unbalanced force working in the horizontal direction.

Sol:

Let v be the velocity of the girl on the cart as the cart starts moving.

The total momenta of the girl and cart before the interaction

= 40 kg \(\times\) 5 ms^–1 + 3 kg \(\times\) 0 ms^–1

= 200 kgms^–1.

Total momenta after the interaction

= (40 + 3) kg \(\times\) v ms^–1

= 43 v kgms^–1.

According to the law of conservation of momentum, the total momentum is conserved during the interaction. That is,

43 v = 200

\(\Rightarrow\) v = 200/43 = + 4.65 ms^–1.

The girl on the cart would move with a velocity of 4.65 ms^–1 in the direction in which the girl jumped (Fig. 9.18)

Figure 9.18(a)(b):  The girl jumps onto the cart

Illustration 9.8:

Two hockey players of opposite teams, while trying to hit a hockey ball on the ground collide and immediately become entangled. One has a mass of 60 kg and was moving with a velocity of 5.0 ms^–1 while the other has a mass of 55 kg and was moving faster with a velocity of 6.0 ms^–1 towards the first player. In which direction and with what velocity will they move after they become entangled? Assume that the frictional force acting between the feet of the two players and the ground is negligible.

Sol:

Let the first player be moving from left to right. By convention left to right is taken as the positive direction and thus right to left is the negative direction (Fig. 9.19). If symbols m and u represent the mass and initial velocity of the two players, respectively. Subscripts 1 and 2 in these physical quantities refer to the two hockey players. Thus, 

m₁  = 60 kg; u₁ = + 5 ms^-1; and

m₂  = 55 kg; u₂ = – 6 ms^-1.

The total momentum of the two players before the collision

= 60 kg \(\times\) (+ 5 ms^-1) + 55 kg \(\times\) (– 6 ms^-1)

=  – 30 kgms^-1

Figure 9.19:  A collision of two hockey players: (a) before the collision and (b) after the collision.

If v is the velocity of the two entangled players after the collision, the total momentum then

=  (m₁ + m₂) \(\times\) v

=  (60 + 55) kg \(\times\) v ms^–1

=   115 \(\times\) v kgms^–1.

Equating the momenta of the system before and after the collision, in accordance with the law of conservation of momentum, we get

v = – 30/115

   = – 0.26 ms^–1.

Thus, the two entangled players would move with velocity 0.26 ms^–1 from right to left, that is, in the direction the second player was moving before the collision.

Questions

1.  If action is always equal to the reaction, explain how a horse can pull a cart.

2.  Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.

3.  From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 ms^–1. Calculate the initial recoil velocity of the rifle.

4.  Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 ms^–1 and 1 ms^–1, respectively. They collide and after the collision, the first object moves at a velocity of 1.67  s^–1. Determine the velocity of the second object.

Source: This topic is taken from NCERT TEXTBOOK

Conservation Laws

CONSERVATION LAWS

All conservation laws such as conservation of momentum, energy, angular momentum, charge, etc. are considered to be fundamental laws in physics. These are based on observations and experiments. It is important to remember that a conservation law cannot be proved. It can be verified, or disproved, by experiments. An experiment whose result is in conformity with the law verifies or substantiates the law; it does not prove the law. On the other hand, a single experiment whose result goes against the law is enough to disprove it.

The law of conservation of momentum has been deduced from a large number of observations and experiments. This law was formulated nearly three centuries ago. It is interesting to note that not a single situation has been realized so far, which contradicts this law. Several experiences of everyday life can be explained on the basis of the law of conservation of momentum.

Source: This topic is taken from NCERT TEXTBOOK