At what speed does a 1200 kg compact car have the same kinetic energy as a 17000 kg truck going 21 km/hr ?

mv² = MV²
1200 * v² = 17000 * (21km/h)²
v = 79 km/h

m1(v1)^2=m2(v2)^2
v1^2=(v2)^2* m2/m1
v1= v2*sqrt(m2/m1)
v1= 21* sqrt(17000/1200)
v1=21 * 3.764 = 79.04

From KE = 1/2 mV^2 = 1/2 Mv^2 = KE, we have V = v sqrt(M/m) = 21*sqrt(170/12) = 70 kph. ANS.

Let v = velocity of car

Then KE of car = 1/2 (1200)v^2

= 600v^2

KE of truck = 1/2(17000)(35/6)^2 .......................21 km/h = 35/6 m/s

= 8500(35/6)^2

Therefore

600v^2 = 8500(35/6)^2

v^2 = (8500/600)(35/6)^2

v = √(8500/600) * 35/6

= 21.95586904

= 21. 96 m/s

K.E. = (1/2) m v²

K.E. of the compact car = K.E. of the truck
(1/2) × (1200 kg) × v² = (1/2) × (17000 kg) × (21 km/hr)²
v² = 17000 × 21² / 1200 (km/hr)²
v = √(1700 × 21² / 1200) km/hr
v = 25 km/hr

Velocity of the compact car = 25 km/hr