**LINEAR MOMENTUM**

Direction of Linear momentum will be just same as direction of velocity

**Case I Changing of linear momentum**

**Note:** (1) Slope of p - t graph will give us Force (F)

(2) Area under the F - t carve will gives us change in momentum

**Impulse **FDt =Dp Also we can say impulse is change in momentum

**Example:** A ball of mass 1 kg drops on the hard floor and ball strikes the floor with 10m/s rebound with same speed also. If ball remains for 0.1 sec in contact with floor. Then find

(i) Magnitude of rate of change of momentum

(ii) Rate of change of magnitude of momentum

**Solution: **Momentum just before strike the floor p_{1} = mv

p_{1} = - 1 x10

= - 10 kg mt/sec

Momentum just after strike the floor

p_{2} = + mv

= + 1 x10

= 10 kg mt/sec

(i) Magnitude of rate change of momentum

** **

(ii) Rate of change of magnitude of momentum

Þ Magnitude of momentum is not changing only direction of momentum is changing

Þ Therefore **Ans-Zero**

**CHANGING IN MOMENTUM**

Change in momentum along *x*- axis Þ

*p*_{1} = *mv *sinq before Collision

*p*_{2 }= *mv* sinq after Collision

(D *p _{x}*) =

*p*

_{2}–

*p*

_{1}= 0

Change in momentum along *y*-axis Þ

*p*_{1} = - *mv* cosq before Collision

*p*_{2} = *mv* cosq after Collision

(Dp)_{y} = p_{2}- p_{1} = 2 mv cosq

**CHANGING IN MOMENTUM **

Let assume a ball attached with a string and moving uniformly in a Circular path

(i) Change in momentum from A to B

(ii) Change in momentum from A to C

i

**Another case**

**But here**

v^{2} = u^{2} – 2gl

**Case I Conservation of linear momentum (COLM)**

**Example:**

Given A trolly of mass M kept at rest on a Smooth road. Two person each of mass m Standing at two ends of trolly. Now person A jumps in horizontal direction form the trolly with speed u (with respect to trolly) and then person B jumps in horizontal direction with speed u (with respect to trolly) in opposite direction. Find final speed of trolly just after both person jumps from the trolly.

**Solution**

Just before A jumps

p_{1} = (2 m + m) x 0 = 0

Just after A jumps

p_{2} = mu + (M + m) v_{1}

Now F_{ext} = 0, applying COLM

p_{1} = p_{2} Þ mu + (M + m) v_{1} = 0

Again if we assume that trolly and person B is at rest

Just before *B* jumps

* p*_{1} = (*M* + *m*) x 0 (Assume)

Just after *B* jumps

* p*_{2} = - *mu* + *Mv*_{2}

Now applying COLM

* p*_{1} = *p*_{2}Þ - *mu* + *Mv*_{2} = 0

Now Final speed of trolly after both A & B Jumps

*V _{f}* =

*V*

_{1}+

*V*

_{2 }

**COLLISION **

Head on Collision

**OBLIQUE COLLISION**

**Types of Collisions **

**(1) Perfect elastic Collision**.

Momentum will Conserved p_{1} = p_{2}

Energy will Conserved k_{1} = k_{2}

No loss of energy

Coefficient of restitution e = 1

**CONCEPT OF EXCHANGE **

**(2) Elastic Collision **

Momentum will Conserved p_{1} = p_{2}

Energy will not Conserved k_{1}¹ k_{2}

Minimum energy loss Coefficient of restitution

e < 1 (But here e will be Close to 1)

**(3) Perfect inelastic Collision**

Momentum will conserved p_{1} = p_{2}

Energy will not conserved k_{1} ≠ k_{2}

Maximum energy loss coefficient of restitution e = 0

**(4) Inelastic Collision **

Momentum will Conserved p_{1} = p_{2}

Energy will not Conserved k_{1} ¹ k_{2}

Maximum energy loss Coefficient of restitution

e > 0 (But here e will be close to 0)

**NEWTONS LAW FOR COLLISION **

Speed just after Collision = e (Speed just before Collision)

Height after Collision = e^{2} (Height before Collision)

Total distance Covered by the ball

H = h + 2 e^{2} h + 2e^{4} h + -------

= h [ 1 + 2e^{2} (1 + e^{2} + e^{4} + ----)]

Continue reading to

Join any of the batches using this book

## Batch List

###### Kaysons Academy

Course : JEE

Start Date : 03.06.2023

End Date : 30.06.2023

Types of Batch : Live Online Class

###### Kaysons Academy

Course : JEE

Start Date : 02.06.2023

End Date : 30.06.2023

Types of Batch : Live Online Class