5.6 Projectile motion on an incline  (Page 7/7)

In the coordinate system of incline and perpendicular to incline, motion parallel to incline denotes a situation when component of velocity in y-direction is zero. Note that this is an analogous situation to the point of maximum height in the normal case when projectile returns to same level.

We shall analyze the situation, taking advantage of this fact. Since component of velocity in y - direction is zero, it means that velocity of projectile is same as that of component velocity in x-direction. For consideration of motion in y -direction, we have :

$$v_y=u_y+a_{y}t$$

$$\Rightarrow 0=30\sin {30}^0-g\cos {30}^{0}Xt$$ $$\Rightarrow t=\frac{15X2}{10X\sqrt{3}}=\sqrt{3}s$$

For consideration of motion in x -direction, we have :

$$v_x=u_x+a_{x}t=30\cos {30}^0-g\sin {30}^{0}Xt$$ $$\Rightarrow v_x=30X\frac{\sqrt{3}}{2}-10X\frac{1}{2}X\sqrt{3}$$ $$\Rightarrow v_x=15X\sqrt{3}-5X\sqrt{3}=10\sqrt{3}$$ $$\Rightarrow v_x=10\sqrt{3}m/s$$

But, component of velocity in y-direction is zero. Hence,

$$v=v_A=10\sqrt{3}m/s$$

Hence, option (d) is correct.

This arrangement is an specific case in which incline plane are right angle to each other. We have actually taken advantage of this fact in assigning our coordinates along the planes, say y-axis along first incline and x-axis against second incline.

The acceleration due to gravity is acting in vertically downward direction. We can get the component accelerations either using the angle of first or second incline. Either of the considerations will yield same result. Considering first incline,

$$a_x=-g\cos {30}^0=-10X\frac{\sqrt{3}}{2}=-5\sqrt{3}m/s^2$$ $$a_y=-g\sin {30}^0=-10X\frac{1}{2}=-5m/s^2$$

Hence, options (a) and (d) are correct.

This arrangement is an specific case in which incline plane are right angle to each other. We have actually taken advantage of this fact in assigning our coordinates along the planes, say y-axis along first incline and x-axis against second incline.

In order to find the time of flight, we can further use the fact that projectile hits the other plane at right angle i.e. parallel to y-axis. This means that component of velocity in x-direction i.e. along the second incline is zero. This, in turn, suggests that we can analyze motion in x-direction to obtain time of flight.

In x-direction,

$$v_x=u_x+a_{x}T$$ $$\Rightarrow 0=u_x+a_{x}T$$ $$\Rightarrow T=-\frac{u_x}{a_x}$$

We know need to know "ux” and "ax” in this coordinate system. The acceleration due to gravity is acting in vertically downward direction. Considering first incline,

$$a_x=-g\cos {30}^0=-10X\frac{\sqrt{3}}{2}=-5\sqrt{3}m/s^2$$ $$a_y=-g\sin {30}^0=-10X\frac{1}{2}=-5m/s^2$$

Also, we observe that projectile is projected at right angle. Hence, component of projection velocity in x - direction is :

$$u_x=10\sqrt{3}m/s$$

Putting values in the equation and solving, we have :

$$\Rightarrow T=-\frac{u_x}{a_x}=\frac{-10\sqrt{3}}{-5\sqrt{3}}=2s$$

Hence, option (b) is correct.

We notice here that initial velocity in y-direction is zero. On the other hand, final velocity in the y-direction is equal to the velocity with which projectile hits at "Q". The x-component of velocity at "Q" is zero. The analysis of motion in y - direction gives us the relation for component of velocity in y-direction as :

In y-direction,

$$v=v_y=u_y+a_y$$ $$\Rightarrow T=0+a_{y}T$$ $$\Rightarrow v=v_y=a_{y}T$$

Thus, we need to know component of acceleration in y-direction and time of flight. As far as components of acceleration are concerned, the acceleration due to gravity is acting in vertically downward direction. Considering first incline, we have :

$$a_x=-g\cos {30}^0=-10X\frac{\sqrt{3}}{2}=-5\sqrt{3}m/s^2$$ $$a_y=-g\sin {30}^0=-10X\frac{1}{2}=-5m/s^2$$

In order to find the time of flight, we can further use the fact that the component of velocity in x-direction i.e. along the second incline is zero. This, in turn, suggests that we can analyze motion in x-direction to obtain time of flight.

In x-direction,

$$v_x=u_x+a_{x}T$$ $$\Rightarrow 0=u_x+a_{x}T$$ $$\Rightarrow T=-\frac{u_x}{a_x}$$

We know need to know "ux" in this coordinate system. We observe that projectile is projected at right angle. Hence, component of projection velocity in x - direction is :

$$u_x=10\sqrt{3}m/s$$

Putting values in the equation and solving, we have :

$$\Rightarrow T=-\frac{u_x}{a_x}=\frac{-10\sqrt{3}}{-5\sqrt{3}}=2s$$

Thus putting values for the expression for the speed of the projectile with which it hits the incline is :

$$\Rightarrow v=v_y=a_{y}T=-5X2=-10m/s$$

Thus, speed is 10 m/s.

Hence, option (b) is correct.