Car A m= 2000kg
v=20 km/h
Car B m= 2000kg
v= 20km/h what is the location after impact

Mass of car, M=2000kg

Velocity, V=72× 5/18 =20ms¹

Apply kinematic equation of motion

v² -u² = as

0 − 20² =2a×20

a=−10ms^-2

Breaking force, F=ma=2000×10=20 kN

Apply first kinematic equation

v = u +at

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electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j