JEE Main Motion in Straight Line PYQ

11. .(JEE Main 2022 (Online) 27th July Evening Shift )

Two inclined planes are placed as shown in figure. A block is projected from the Point A of inclined plane AB along its surface with a velocity just sufficient to carry it to the top Point B at a height 10 m. After reaching the Point B the block slides down on inclined plane BC. Time it takes to reach to the point C from point A is $t (\sqrt{2} + 1)$ s. The value of t is ___________.

(use $g = 10 m / s^{2}$ )

JEE Main 2022 (Online) 27th July Evening Shift Physics - Motion Question 42 English

Correct answer is (2)

$A B = 10 \sqrt{2}$ m

$v_{A} = \sqrt{2 \times 10 \times 10} = 10 \sqrt{2}$ m/s

$v_{C} = 10 \sqrt{2}$ m/s

$a_{B C} = g \sin (30^{\circ}) = 5$ m/s²

$t_{B C} = 2 \sqrt{2} s (\frac{v_{c}}{a_{B C}})$

$t_{A B} = \frac{v_{A}}{5 \sqrt{2}} = 2 s$

$t_{A B} + t_{B C} = 2 (\sqrt{2} + 1)$

$\Rightarrow t = 2$

12. .(JEE Main 2022 (Online) 27th July Morning Shift )

A ball of mass m is thrown vertically upward. Another ball of mass $2 m$ is thrown at an angle $θ$ with the vertical. Both the balls stay in air for the same period of time. The ratio of the heights attained by the two balls respectively is $\frac{1}{x}$ . The value of x is _____________.

Correct answer is (1)

JEE Main 2022 (Online) 27th July Morning Shift Physics - Motion Question 44 English Explanation

$∴$ $u_{1} = u_{2} \sin θ$

$\frac{H_{1}}{H_{2}} = \frac{\frac{u_{1}^{2}}{2 g}}{u_{2}^{2} \frac{\sin^{2} θ}{2 g}}$

$= {(\frac{u_{1}}{u_{2} \sin θ})}^{2} = 1$

13. .(JEE Main 2022 (Online) 26th June Evening Shift )

A ball is projected vertically upward with an initial velocity of 50 ms^{$-$ 1} at t = 0s. At t = 2s, another ball is projected vertically upward with same velocity. At t = __________ s, second ball will meet the first ball (g = 10 ms^{$-$ 2}).

Correct answer is (6)

At t = 2 s, v₁ = 50 $-$ 2 $\times$ 10 = 30 m/s

v₂ = v₂

$∴$ a_rel = g $-$ g = 0

$S = \frac{u^{2} - v^{2}}{2 g} = \frac{50^{2} - 30^{2}}{2 \times 10} = \frac{1600}{20} = 80$ m

$∴$ v_rel = 50 $-$ 30 = 20 m/s

$∴$ $Δ t = \frac{80}{20} = 4 s$

$∴$ required time t = 2 + 4 = 6 s

14. .(JEE Main 2022 (Online) 26th June Morning Shift )

A ball of mass 0.5 kg is dropped from the height of 10 m. The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is ________ m. [Use g = 10 m/s²]

Correct answer is (5)

JEE Main 2022 (Online) 26th June Morning Shift Physics - Motion Question 60 English Explanation

Let at h height its velocity becomes 10 m/s (as given in question)

$\Rightarrow v^{2} - u^{2} = 2 (- 10) \times (- h)$

$\Rightarrow v^{2} = 20 h$

$\Rightarrow 100 = 20 h$

$\Rightarrow h = 5 m$

$h^{'} = 10 - 5 = 5 m$

15. .(JEE Main 2022 (Online) 24th June Morning Shift )

From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6 s. A second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 s. A third ball released, from the rest from the same location, will reach the ground in ____________ s.

Correct answer is (3)

Based on the situation

$h = - u t_{1} + \frac{1}{2} g t_{1}^{2}$ $\to$ throwing up ....... (i)

$h = u t_{2} + \frac{1}{2} g t_{2}^{2}$ $\to$ throwing up ....... (ii)

$h = \frac{1}{2} g t^{2}$ $\to$ dropping .......... (iii)

and $0 = u (t_{1} - t_{2}) - \frac{1}{2} g (t_{1} - t_{2})^{2}$ ....... (iv)

solving above equations

$t = \sqrt{t_{1} t_{2}}$

$\Rightarrow t = \sqrt{6 \times 1.5} = 3 s$