question_answer

A body of mass M at rest explodes into three pieces, two of them of mass M/4 each, are thrown off in perpendicular directions with velocities of 3 m/s and 4 m/s respectively. The third piece will be thrown off with a velocity of

A) 1.5 m/s

B)  2 m/s 

C)  2.5 m/s

D)  3 m/s

Correct Answer: C

Solution :

 Let one mass piece of mass M/4 be thrown off in the horizontal direction with a speed of 3m/s, and the other mass piece of mass M/4 be thrown in the vertical direction with a speed of 4m/s and the remaining mass M/2 be thrown with a velocity vm/s, making an angle \[\theta \] with the horizontal direction. So, from conservation of momentum, Along horizontal direction \[\frac{M}{4}\times 3+\frac{M}{2}\times v\cos \theta =0\] ... (i) Along vertical direction \[\frac{M}{4}\times 4+\frac{M}{2}\times v\sin \theta =0\] ... (ii) So, from Eqs. (i) and (ii), we get \[v\cos \theta =-\frac{3}{2}\] ... (iii) and \[v\sin \theta =-2\] ?. (iv) So, \[v=\sqrt{{{(3/2)}^{2}}+{{(2)}^{2}}}\] \[=\sqrt{9/4+4}=\frac{5}{2}\]