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Basic Numericals based on Collision Part -3 (In Hindi)
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Through this lesson, you can learn how to find momentum, impulse (by formula), force, energy loss, impulse (graphical condition) etc.

## Kumar Ketan is teaching live on Unacademy Plus

Kumar Ketan
#Specialisation in Electronics #6 years teaching experience #Physics Faculty in Kota #"magical education" (211k+ subs) #Writer

U
thank you
Sir according to integration, I = 1/2 F × t^2/2. but it is not so . y?
Kumar Ketan
8 months ago
Pls mention the time of doubt
dhasuu .............sir ji
Kumar Ketan
9 months ago
Ty beta ji
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2. Problems Of Collision 2

3. Example: A 1500 kg car collides with a wall, with V -15m/s and v 2.6m/s. What is the impulse exerted on the car? m (f = (1500kg =26400 NS Before 2.6 m /S -15.0 m/s If the collision time is 0.15 s, what is the average force exerted on the car? After F): 26400 Ns 0.15 s = 1.76 10 5 N - +2.60 m/s Energy loss due to collision

4. Example of graphical representation of F(t) F (N) max = 18 000 N 20 000 15 O00 10 000 5 000 t (ms) 0 2 Estimate of impulse: 0.0015 != F (t)dt =-Fmax (0.0015 s)--(18000 N)(0.0015 s)= 13.5 Ns

5. Example - Completely inelastic collision; balls moving on a frictionless surface initial P final Before the collision, the particles move separately 22i vii mo For m1=0.3kg,m2=0.5kg After the collision, the particles move together. (0.3 X2)i + (0.5X-1 (0.3)+ (0.5 m S mi m2 0.125 m/si

6. Energy loss in this example: Before the collision, the particles move separately For m0.3kg, m 0.5kg m1 2 = 0.125 m/s ! AE -((0.3)+ (0.5))0.125 0.3 2(0.51 After the collision, the particles move together. =-084 J m1 m2

7. Example- completely elastic collision, balls moving on a frictionless surface i initial i final Before the collision, the particles move separately m,v i initial ii final 2 2i 2 2 For 1- d motion, after some algebra: After the collision, the particles continue to move separately with new velocities. im 2 m 2 1 m1 t m 2 V2f 1m 1m m1tn

8. Completely elastic collision; numerical example: 2 m 2 t Before the collision, the particles move separately. mi+ mm mi + m 2 2 m VLE t Vii For m0.3kg 0.5 kg 2 2 i (03)-(03) ( 2 m s+ 0.3)+ (0.5 2 (0.5 (0.3)+0.5 2 (0.3 0.3(0.5 After the collision, the particles continue to move separately with new velocities. 1m/s)=-1.75 ml s 0.3)+ (0.5