A projectile of mass ` 100 g` is fired with a velocity of ` 20 ms^(-1)` making an angle of ` 30^@` with the borizontal. As it rises to the highest point of its path its momentum changes by .

Options

A. ` 1//2 kg ms^(-1)`
B. `1 kg ms^(-1)`
C. ` 2 kg ms^(-1)`
D. none of these

Answer

B

Explanation

Ist method. Initial momentum of the body,
` vec P_1 = m vec u = (0.1 xx 20 ) ` along ` OA`
Final momentme, ` vec P_2 = mu cos hteta` along ` OB`
` =0.1 xx 20 xx cos 30^@`
` = 0.1 xx 20 xx (sqrt 3)/2 = sqrt 3 kg ms^(-1)` along ` OB`.
.
Change in momentum
` = vec P_2 - vec P-1 vec (OB) - vec (OA) = vec (AB)`
` Now , ` | vec AB| = sqrt ( (OA)^2 - (OB)^2)`
` = sqrt (2^2 - ( sqrt 3)^@) = 1 kg ms^(-1)`.
2nd methok. Horizontal momentum of the body at (O) and at highest poin is same . The vertical momenitum at ` O = mv sin theta ` and at hifhest point is ` zero . Therefore, charge in momentum`
` = mv sin hteta = 0.1 xx 20 xx sin 30^@ = 1 kg ms^(-1)`.