Question

The small marble is projected with a velocity of $10 \,ms ^{-1}$ in a direction $45^{\circ}$ from the horizontal $y$-direction on the smooth inclined plane. Calculate the magnitude $v$ of its velocity after $2 s$.

• A. $10 \sqrt{2} m s ^{-1}$
• B. $5\, ms ^{-1}$
• C. $10\, ms ^{-1}$
• D. $5 \sqrt{2} m s ^{-1}$

Answer 1

Answer: (C)

Given $\left(V=10\, ms ^{-1}\right)$
After 2s: $V_{x}=\frac{10}{\sqrt{2}}-\frac{10}{\sqrt{2}} \times 2$
$\Rightarrow V_{x}=-\frac{10}{\sqrt{2}} m s ^{-1}$
$V_{x}=-\frac{10}{\sqrt{2}} m s ^{-1} \text { and } V_{y}=-\frac{10}{\sqrt{2}} m s ^{-1}$
$V=\sqrt{\frac{100}{2}+\frac{100}{2}}=10 \sqrt{\frac{1}{2}+\frac{1}{2}}$
$=10 m s ^{-1}$