Question

lorry and a ca of mass ratio 4;1 are moving with ke in the ratio 3;2 on a horizontal road
now breaks are applied and breaking forces are in the ratio 1;2 then the stopping times of lorry and car in the ratio

Answer 1

Let mass of car is m
∴ mass of lorry is 4m
Kinetic energy of car is 2K.E
∴ kinetic energy of lorry is 3K.E

∵ kinetic energy = 1/2 mv²
∴ kinetic energy of car/kinetic energy of lorry = 1/2 mu₁²/1/2 4mu₂²
2/3 = u₁²/4u₂²
⇒u₁²/u₂² = 8/3
⇒u₁/u₂ = √(8/3)

And force = mass × acceleration
So, force applied on car/force applied on lorry = ma₁ /4ma₂
⇒ 2/1 = a₁/4a₂
⇒a₁/a₂ = 8/1

Now, use formula,
v = u + at for both cases,
For car,
v₁ = u₁ + a₁t₁
t₁ = u₁/a₁ [ here acceleration is negative because force applied opposite direction for stopping ]
Similarly for lorry, t₂ = u₂/a₂

Now, t₂/t₁ = u₂a₁/u₁a₂ = (u₂/u₁) × (a₁/a₂)
= √(3/8) × 8
= √{24}

Hence, answer is √24:1