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)mv^{2}×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) Ft^{2}/2m **

**(b) F^{2}t/2m **

**(c) F^{2}t^{2}/2m **

**(d) F^{2}t^{2}/3m **

**(e) F^{2}t^{2}/m **

The impulse received by the particle during the time *t *is *Ft*.*Ft* is the final momentum of the particle. The kinetic energy of the particle after the time *t* is therefore equal to *F*^{2}*t*^{2}/2*m* (remembering that KE = *p*^{2}/2*m* where *p* is the momentum).** **