Two Multiple Choice Questions involving Kinetic Energy

Check whether you can find the answers to the following two questions in a couple of minutes:

(1) When a running boy increases his speed by 2 ms^–1, his kinetic energy becomes three times the original value. His original speed is (in ms^–1)

(a) √3 +1

(b) √3 –1

(c) √3

(d) 2√3

(e) 2

If his original speed is ‘v’ we can write

(½)mv²×3 = (½)m(v+2)² where ‘m’ is his mass.

This gives v√3 = (v+2) so that v(√3 –1) =2.

Therefore, v = 2 /(√3 –1) = √3 +1 on multiplying the numerator and denominator by (√3 –1).

(2) A particle of mass ‘m’ at rest is acted on by a force ‘F’ for a time‘t’. Its kinetic energy after the time t is

(a) Ft²/2m

(b) F²t/2m

(c) F²t²/2m

(d) F²t²/3m

(e) F²t²/m

The impulse received by the particle during the time t is Ft. But impulse is equal to the change of momentum. Since the initial momentum of the particle is zero, Ft is the final momentum of the particle. The kinetic energy of the particle after the time t is therefore equal to F²t²/2m (remembering that KE = p²/2m where p is the momentum).