A particle A moves along the line y=30 m wit | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the time of collision be $t$. At the time of collision, both particles must have the same $x$ and $y$ coordinates.For particle $A$:$x_A = v t$

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Let the time of collision be $t$. At the time of collision, both particles must have the same $x$ and $y$ coordinates.For particle $A$:$x_A = v t$, $y_A = 30 \ m$.For particle $B$:$a_x = a \sin 60^{\circ} = 0.4 	imes \frac{\sqrt{3}}{2} = 0.2\sqrt{3} \ m/s^2$.$a_y = a \cos 60^{\circ} = 0.4 	imes \frac{1}{2} = 0.2 \ m/s^2$.Since $B$ starts from the origin with zero initial velocity:$x_B = \frac{1}{2} a_x t^2 = \frac{1}{2} (0.2\sqrt{3}) t^2 = 0.1\sqrt{3} t^2$.$y_B = \frac{1}{2} a_y t^2 = \frac{1}{2} (0.2) t^2 = 0.1 t^2$.For collision, $y_A = y_B$:$30 = 0.1 t^2 \Rightarrow t^2 = 300 \Rightarrow t = 10\sqrt{3} \ s$.For collision, $x_A = x_B$:$v t = 0.1\sqrt{3} t^2 \Rightarrow v = 0.1\sqrt{3} t$.Substituting $t = 10\sqrt{3}$:$v = 0.1\sqrt{3} 	imes 10\sqrt{3} = 1 	imes 3 = 3 \ m/s$.Thus, the value of $|v|$ is $3 \ m/s$.

Similar Questions

$A$ particle of mass $m$ is projected with velocity $v$ making an angle of $45^\circ$ with the horizontal. When the particle lands on the level ground,the magnitude of the change in its momentum will be

Two particles move on a circular path (one just inside and the other just outside) with angular velocities $\omega$ and $5\omega$ starting from the same point. Then

The displacement vector of a particle of mass $m$ is given by $\vec{r}(t) = A \cos \omega t \hat{i} + B \sin \omega t \hat{j}$.
$(a)$ Show that the trajectory is an ellipse.
$(b)$ Show that $\vec{F} = -m \omega^2 \vec{r}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module