Question

A ball is moving with velocity 5 m/s in a direction which makes an angle of 30° with horizontal (i.e. with positive x-direction). This ball hits a vertical wall and gets reflected in a direction which makes an angle of 60° with the vertical (i.e. with positive y-direction) and moves with the same magnitude i.e. 5 m/s:
a. Find components of the initial and reflected velocities.
b. Show the vector representing the change in velocity of the ball.
c. Find the magnitude of the change in velocity.

Answer 1

Answer:

Explanation:

(a)

From the given information:

The initial velocity $v_1$ = 5 m/s

The direction of the angle θ = 30°

Therefore, the component along the x-axis = $v_1 \ cos \ \theta$

$v_{1 \ x } = 5 \ cos \ 30^0$

$v_{1x} = 4.33 \ m/s$

The component along the y-axis = $v_2 { \ sin \ \theta}$

$v_{1 \ y } = 5 \ sin \ 30^0$

$v_{1 \ y } = 2.5 \ m/s$

To find the final velocity( reflected velocity)

using the same magnitude $v_2 = 5 \ m/s$

The angle from the x-axis can be $\theta_r = 90^0+60^0$

= 150°

Thus, the component along the x-axis = $v_2 \ cos \theta _r$

$v_{2x} = - 0.433 \ m/s$

The component along the y-axis = $v_2 \ sin \theta_r$

$v_{2y} = 5 \ sin \ 150^0$

$v_{2y} = 2.5 \ m/s$

(b)

The velocity $v_1$ can be written as in vector form.

$v_1 ^{\to} = v_1 x \hat {i} + v_1 y \hat {j}$

$v_1 ^{\to} =4.33 \ \hat {i} + 2.5 \ \hat {j}$  ----  (1)

The reflected velocity in vector form can be computed as:

$v_2 ^{\to} = v_2 x \hat {i} + v_2 y \hat {j}$

$v_2 ^{\to} =-4.33 \ \hat {i} + 2.5 \ \hat {j}$  --- (2)

The change in velocity = $v_2 ^{\to} - v_1 ^{\to}$

$\Delta v ^{\to} = - 4.33 \hat i + 2.5 \hat j - 4.33 \hat i - 2.5 \hat j$

$\Delta v ^{\to} = - 8.66 \hat { i }$

(c)

The magnitude of change in velocity = $| \Delta V |$

$| \Delta V |$ = 8.66 m/s