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Two vehicles moving on a horizontal road are stopped by same retarding force.

Column  i Column  ii
(a) When they have same K.E. (P) faster body stop in larger distance
(b) When they have different masses but same velocity (Q) larger body stops in larger distance.
(c) When both have same momentum (R) heavier body stops in lighter distance.
(d) When both have same mass but different velocities (S) stopped in same distance.

(A) a – P, b – Q, c – R, d – S 

(B) a – Q, b – P, c – S, d – R 

(C) a – S, b – R, c – Q, d – P 

(D) a – R, b – S, c – Q, d – P

1 Answer

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Best answer

Correct option: (C) a – S, b – R, c – Q, d – p

Explanation:

F × x = KE

i.e. stopping distance = x = [{KE(E)} / {stopping force (F)}]

x = {(mv2) / (2F)}

Given – retarding force in same

Let m1, m2 be masses of vehicles moving with velocity v1, v2.

x ∝ mv2 

(x1 / x2) = (E1 / E2) = {(m1v12) / (m2v22)}

(1) if KE is same E1 = E2 then

x1 = x2 

hence both will stop at same distance

(2) if velocity is same

(x1 / x2) = (m1 / m2)

i.e. x ∝ m

hence heavy body will take mare distance to stop (3) same mass means

(x1 / x2) = (v1 / v2)2 

 i.e. x ∝ v2 

hence faster body will cover large distance to stop.

(4) if same momentum

KE = (P2 / 2m)

(x1 / x2) = (E1 / E2) = {(P12) / (2m1)} ∙ {(2m2) / (P22)}

P1 = P2 hence

(x1 / x2) = (m2 / m1)

i.e. lighter body stops in large distances heavier stops in small distance.

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