Question

A particle of mass 100 g is fired with a velocity 20 msec^–1 making an angle of 30^o with the horizontal. When it rises to the highest point of its path then the change in its momentum is

• A. $\sqrt{3}kgm\sec^{- 1}{}$
• B. 1/2 kgmsec^–1
• C. $\sqrt{2}kgm\sec^{- 1}{}$
• D. 1 kgmsec^–1

Answer 1

Answer: (D)

1 kgmsec^–1

Answer 2

Horizontal momentum remains always constant

So change in vertical momentum (∆$\overrightarrow{p}$) = Final vertical momentum – Initial vertical momentum $= 0 - mu\sin\theta$

$|\Delta P| = 0.1 \times 20 \times \sin 30^{o}$ $= 1kgm/sec$.