Question

A particle moving with speed 10√2 m/s strikes a wall at an angle 45degree as shown in figure. Find magnitude of change in velocity.

Answer 1

Answer:

No change

Explanation:

Initially ,

Vx1 = V1sin 45° = V1/√2 = 10

Vy1 = V1cos 45° = V1/√2 = 10

Finally,

Vx2 = V2sin 45° = V1/√2 = 10

Vy2 = V2cos 45° = V1/√2 = 10

So, no change in magnitude .

Answer 2

Answer:

The magnitude of change in velocity is 20m/s.

Explanation:

The initial velocity of the ball is given as (v₁)= $v_{0} \cos 45^{\circ} \uparrow+v_{0} \sin 45^{\circ} \hat{\jmath}$

$v_1=\frac{v_{0}}{\sqrt{2}} \hat{\imath}+\frac{v_{0}}{\sqrt{2}} \hat{\jmath}$    (1)

The final velocity is (v₂) $=-v_{0} \cos 45 \hat{\imath}+v_{0} \sin 45 \hat{\jmath}$

$v_2=-\frac{v_{0}}{\sqrt{2}} \hat{i}+\frac{v_{0}}{v_{2}} \hat{\jmath}$      (2)

And the change in velocity is calculated as,

$\Delta v=v_{2}-v_{1}$      (3)

When we enter the values into equation (3), we obtain;

$\Delta v=\left(\frac{-v_{0}}{v_{2}} \hat{\jmath}+\frac{v_{0}}{y_{2}} j\right)-\left(\frac{v_{0}}{v_{2}} j+\frac{v_{0}}{\sqrt{2}} \mathfrak{v}\right)$

$\Delta v=-\frac{v_{0}}{\sqrt{2}} \times 2 \hat{\jmath}=-\sqrt{2} v_{0} \hat{j}$

$|\Delta v|= \sqrt{2} \times 10\sqrt{2} =20\ m/s$     (In question $v_o=10\sqrt{2}$)

The magnitude of change in velocity is 20m/s.

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