Answer to Question #97421 in Physics for Sidhardha

Question

Two inctined plaues OA and OB having
inclination 30 and 60° with the horizontal
respectively intersects each ofher at 0, as
shown in ligure. A particle is projected Irom
point P with a velocity u-103m/s alonga
direction perpendicular to plane OA. If the
particle strikes plane OB perpendicular at

Which of the following is/are correct

A) The tine ol llight 2s
B) The velocity with which the particle
strikes thec planc OB - 10m/s
C) The height of the point P Irou point O is
8m
D) The distance PO-20m

Answer 1

Let's choose the x and y directions along OB and OA respectively. Then,

$$u_x=u=10\sqrt{3}\frac{m}{s},u_y=0$$

$$a_x=-g\sin{60}=-5\sqrt{3}\frac{m}{s^2}$$

$$a_y=-g\cos{60}=-5\frac{m}{s^2}$$

a)  At point Q, x-component of velocity is zero. Hence, substituting in

$$v_x=u_x+a_xt\to 0=10\sqrt{3}-5\sqrt{3}t$$

$$t=2\ s$$

b)  At point Q,

$$v=v_y=u_y+a_yt$$

$$v=0-5(2)=-10\frac{m}{s}$$

c)

$$s_y-PO=v_yt+0.5a_yt^2=0-0.5(5)(2)^2=-10\ m$$

$$h=|s_y|\sin{30}=10(0.5)=5\ m$$

d)

$$s_x=v_xt+0.5a_xt^2=(10\sqrt{3})(2)-0.5(5\sqrt{3})(2)^2=10\sqrt{3}\ m$$

$$PQ=\sqrt{PO^2+QO^2}=\sqrt{10^2+(10\sqrt{3})^2}=20\ m$$

Correct answers: A, B