A 3200 kg police car traveling North at 40 m/s collides with another car also traveling North. The second car has a mass of 2000 kg and its velocity is 30 m/s North just before the collision. If the cars stick together on contact what will be their common velocity just after collision? Neglect the effect of friction from the road.

Explanation:

we are to use the law of momentum

where by total momentum before collision is equal to total momentum after collision hence m1u1+m2u2 =V(M1+M2)

(3200×40)+(2000×30)=V(3200+2000)

128000+60000=V(5200)

188000=5200V

188000÷5200=5200V÷5200

V=36.154m/s

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray

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