Please Wait... While Loading Full Video### CONSERVATION OF MOMENTUM

`color{red} ♦` 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_(AB)` 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)].

From Eq. (9.1), the momenta (plural of momentum) of ball A before and after the collision are `m_A u_A` and `m_A v_A`, respectively. The rate of change of its momentum (or `F_(AB)`, action) during the collision will be ` m_A ( v_A - u_A)/t`

Similarly, the rate of change of momentum of ball `B` (`= F_(BA)` or reaction) during the collision will be ` m_B ( v_B - u_B)/t`

According to the third law of motion, the force `F_(AB)` exerted by ball A on ball B (action) and the force `F_(BA)` exerted by the ball B on ball A (reaction) must be equal and opposite to each other. Therefore,

`F_(AB) = – F_(BA)`(9.6)

or ` m_A ( v_A - u_A)/t = m_B ( v_B - u_B)/t`

This gives,

`m_A u_A + m_B u_B = m_A v_A + m_B v_B` (9.7)

Since `(m_A u_A + m_B u_B)` is the total momentum of the two balls A and B before the collision and `(m_A v_A + m_B v_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 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.

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_(AB)` 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)].

From Eq. (9.1), the momenta (plural of momentum) of ball A before and after the collision are `m_A u_A` and `m_A v_A`, respectively. The rate of change of its momentum (or `F_(AB)`, action) during the collision will be ` m_A ( v_A - u_A)/t`

Similarly, the rate of change of momentum of ball `B` (`= F_(BA)` or reaction) during the collision will be ` m_B ( v_B - u_B)/t`

According to the third law of motion, the force `F_(AB)` exerted by ball A on ball B (action) and the force `F_(BA)` exerted by the ball B on ball A (reaction) must be equal and opposite to each other. Therefore,

`F_(AB) = – F_(BA)`(9.6)

or ` m_A ( v_A - u_A)/t = m_B ( v_B - u_B)/t`

This gives,

`m_A u_A + m_B u_B = m_A v_A + m_B v_B` (9.7)

Since `(m_A u_A + m_B u_B)` is the total momentum of the two balls A and B before the collision and `(m_A v_A + m_B v_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 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.

Activity ______________ `9.5`

♦ 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 balloon. Let the air escape from the mouth of the balloon.

♦ Observe the direction in which the straw moves.

Activity ______________ `9.6`

♦ 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.

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

Q 3255101064

A bullet of mass `20 g` is horizontally fired with a velocity `150 m \ \ s^(-1)` from a pistol of mass `2` kg. What is the recoil velocity of the pistol?

Class 9 Chapter 9 Example 6

Class 9 Chapter 9 Example 6

We have the mass of bullet,

`m_1 = 20 g (= 0.02 kg)` and the mass of

the pistol, `m_2 = 2` kg; initial velocities of

the bullet `(u_1)` and pistol `(u_2) = 0`,

respectively. The final velocity of the

bullet, `v_1 = + 150 m \ \ s^(-1)`. The direction

of bullet is taken from left to right

(positive, by convention, Fig. 9.17). 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 xx 0 m s^(–1)`

`= 0 kg m \ \ s^(–1)`

Total momenta of the pistol and bullet after it is fired

`= 0.02 kg xx (+ 150 m \ \ s^(–1) )`

`+ 2 kg xx v m \ \ s^(–1)`

`= (3 + 2 v) kg m \ \ s^(–1)`

According to the law of conservation of momentum

Total momenta after the fire = Total momenta before the fire

`3 + 2v = 0`

`⇒ v = − 1.5 m \ \ s^(–1)`.

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

Q 3265101065

A girl of mass `40` kg jumps with a horizontal velocity of `5 m \ \ s^(-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.

Class 9 Chapter 9 Example 7

Class 9 Chapter 9 Example 7

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 xx 5 m \ \ s^(–1) + 3 kg xx 0 m \ \ s^(–1)`

`= 200 kg m \ \ s^(–1)`.

Total momenta after the interaction

`= (40 + 3) kg xx v m \ \ s^(–1)`

`= 43 v kg m \ \ s^(–1)`.

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

That is,

`43 v = 200`

`⇒ v = 200//43 = + 4.65 m s^(–1)`.

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

Q 3275101066

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 `5.0 m \ \ s^(–1)` while the other has a mass of 55 kg and was moving faster with a velocity `6.0 m \ \ s^(–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 ground is negligible.

Class 9 Chapter 9 Example 8

Class 9 Chapter 9 Example 8

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

`= (m_1 + m_2) xx v`

`= (60 + 55) kg xx v m \ \ s^(–1)`

`= 115 xx v kg m \ \ s^(–1)`.

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

`v = – 30//115`

`= – 0.26 m \ \ s^(–1)`.

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

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,

`m_1 = 60 kg; u_1 = + 5 m \ \ s^(-1) `; and

`m_2 = 55 kg; u_2 = – 6 m \ \ s^(-1)`.

The total momentum of the two players before the collision

`= 60 kg xx (+ 5 m \ \ s^(-1) ) + 55 k g xx (– 6 m \ \ s^(-1) )`

`= – 30 kg m \ \ s^(-1)`