Question

An object of mass 3 kg is moving with a velocity of 5 m/s along a straight path. If a force of 12 N is applied for 3 sec on the object in a perpendicular to its direction of motion. The magnitude of velocity of the particle at the end of 3 sec is _______ m/s. 

(A) 5 

(B) 12 

(C) 13 

(D) 4

Answer 1

Correct option: (C) 13

Explanation:

Given : m = 3 kg 

v = 5 k/s 

F = 12 N 

t = 3 sec

F × t = m₁ v₁ – m₂ v₂ 

for the vertically applied force,

12 × 3 = 3(v₁ – v₂)

12 = v₁ – v₂ 

∴ v₁ = 0

 ∴ v₂ = – 12

v₂ = 12 m/s acting in down ward direction, hence resultant velocity is

V = √(12² + 5²)

V = 13 m/s