Question

The force on a particle of mass $10 \, g$ is $\left(10 \, \hat{i } + 5 \, \hat{j ⁡}\right)\text{N}$ . If it starts from rest at the origin, what would be its position at time $t=5 \, s$ ?

• A. $\text{12500}\hat{i }+\text{6250}\hat{j ⁡}\text{ m}$
• B. $\text{6250}\hat{i }+\text{12500}\hat{j ⁡}\text{ m}$
• C. $\text{12500}\hat{i }+\text{12500}\hat{j ⁡}\text{ m}$
• D. $\text{6250}\hat{i }+\text{6250}\hat{j ⁡}\text{ m}$

Answer 1

Answer: (A)

From $\overset{ \rightarrow }{F }=\left(10 \hat{i ⁡} + 5 \hat{j ⁡}\right)\text{N}$ , we have
$ F _{ x ⁡} = 1 0 \text{ N} \text{; } F ⁡_{ y ⁡} = 5 \text{ N}$
$a _{x ⁡}=\frac{F ⁡_{x ⁡}}{m ⁡}=\frac{1 0}{\text{0.01}}=1000\text{ m s}^{- 2}$
As acceleration along the x-axis is constant,
$\therefore x = u ⁡_{ x ⁡} t ⁡ + \frac{1}{2} a ⁡_{ x ⁡} t ⁡^{2}$
$= 0 + \frac{1}{2} \times 1 0 0 0 \times 5^{2} = 1 2 5 0 0 \text{ m}$
Similarly, $a _{y ⁡}=\frac{F ⁡_{y ⁡}}{m ⁡}=\frac{5}{\text{0.01}}=500\text{ m s}^{- 2}$
and $ y = u ⁡_{ y ⁡} t ⁡ + \frac{1}{2} a ⁡_{ y ⁡} t ⁡^{2} = 0 + \frac{1}{2} \times 5 0 0 \times 5^{2} = 6 2 5 0 \text{ m}$
Hence the position of the particle at $t=5s$ is
$\overset{ \rightarrow }{r }=\left(12500 \hat{i ⁡} + 6250 \hat{j ⁡}\right)\text{m}$