Top 50 Most Repeated MOTION IN A PLANE PYQs | JEE MAINS
A curated collection of the most important questions from MOTION IN A PLANE, fully solved with step-by-step concepts to prepare for JEE MAINS.
A curated collection of the most important questions from MOTION IN A PLANE, fully solved with step-by-step concepts to prepare for JEE MAINS.
$H = \frac{u^2 \sin^2\theta}{2g} = \frac{400 \cdot (\sqrt{3}/2)^2}{20} = \frac{400 \cdot 0.75}{20} = 15\,\text{m}$...
Read Full Step-by-Step Solution →Horizontal component relative to ground: \(v_{x}=20\cos30^{\circ}+10 = 17.32+10 = 27.32\) m/s. Vertical component: \(v_{y}=20\sin30^{\circ}=10\) m/s. ...
Read Full Step-by-Step Solution →Relative velocity vector: $\vec{v}_{bg} = \vec{v}_{br} + \vec{v}_{rg}$. Magnitude = $\sqrt{3^2 + 4^2} = 5\,\text{km/h}$....
Read Full Step-by-Step Solution →Horizontal component relative to the platform: $u\cos30^{\circ}=20\times\frac{\sqrt3}{2}=17.32\,\text{m s}^{-1}$. Adding platform speed gives
Using \(\omega = \omega_0 + \alpha t\) with \(\omega_0 = 0\), \(\alpha = 2\) rad·s⁻² and \(t = 3\) s, we get \(\omega = 0 + 2\times3 = 6\) rad·s⁻¹....
Read Full Step-by-Step Solution →Vertical component: \(u_y = 30\sin30^{\circ}=15\) m/s, so time of flight \(T = 2u_y/g \approx 3.06\) s. Horizontal component relative to train: \(u_x ...
Read Full Step-by-Step Solution →The horizontal component of velocity remains constant, while the vertical component becomes zero at the maximum height....
Read Full Step-by-Step Solution →To find the horizontal distance covered, we need to find the horizontal component of the initial velocity and then use it to calculate the distance. T...
Read Full Step-by-Step Solution →The change in momentum is equal to the initial momentum, which is the product of the mass and velocity of the particle....
Read Full Step-by-Step Solution →At the highest point, the vertical component of velocity is 0. The initial vertical component of velocity is 50 sin(60°) = 25√3 m/s. Since the acceler...
Read Full Step-by-Step Solution →$H = \frac{u^2 \sin^2\theta}{2g} = \frac{400 \cdot (\sqrt{3}/2)^2}{20} = \frac{400 \cdot 0.75}{20} = 15\,\text{m}$...
Read Full Step-by-Step Solution →Horizontal component relative to ground: \(v_{x}=20\cos30^{\circ}+10 = 17.32+10 = 27.32\) m/s. Vertical component: \(v_{y}=20\sin30^{\circ}=10\) m/s. ...
Read Full Step-by-Step Solution →Relative velocity vector: $\vec{v}_{bg} = \vec{v}_{br} + \vec{v}_{rg}$. Magnitude = $\sqrt{3^2 + 4^2} = 5\,\text{km/h}$....
Read Full Step-by-Step Solution →Horizontal component relative to the platform: $u\cos30^{\circ}=20\times\frac{\sqrt3}{2}=17.32\,\text{m s}^{-1}$. Adding platform speed gives
Using \(\omega = \omega_0 + \alpha t\) with \(\omega_0 = 0\), \(\alpha = 2\) rad·s⁻² and \(t = 3\) s, we get \(\omega = 0 + 2\times3 = 6\) rad·s⁻¹....
Read Full Step-by-Step Solution →Vertical component: \(u_y = 30\sin30^{\circ}=15\) m/s, so time of flight \(T = 2u_y/g \approx 3.06\) s. Horizontal component relative to train: \(u_x ...
Read Full Step-by-Step Solution →The horizontal component of velocity remains constant, while the vertical component becomes zero at the maximum height....
Read Full Step-by-Step Solution →To find the horizontal distance covered, we need to find the horizontal component of the initial velocity and then use it to calculate the distance. T...
Read Full Step-by-Step Solution →The change in momentum is equal to the initial momentum, which is the product of the mass and velocity of the particle....
Read Full Step-by-Step Solution →At the highest point, the vertical component of velocity is 0. The initial vertical component of velocity is 50 sin(60°) = 25√3 m/s. Since the acceler...
Read Full Step-by-Step Solution →$H = \frac{u^2 \sin^2\theta}{2g} = \frac{400 \cdot (\sqrt{3}/2)^2}{20} = \frac{400 \cdot 0.75}{20} = 15\,\text{m}$...
Read Full Step-by-Step Solution →Horizontal component relative to ground: \(v_{x}=20\cos30^{\circ}+10 = 17.32+10 = 27.32\) m/s. Vertical component: \(v_{y}=20\sin30^{\circ}=10\) m/s. ...
Read Full Step-by-Step Solution →Relative velocity vector: $\vec{v}_{bg} = \vec{v}_{br} + \vec{v}_{rg}$. Magnitude = $\sqrt{3^2 + 4^2} = 5\,\text{km/h}$....
Read Full Step-by-Step Solution →Horizontal component relative to the platform: $u\cos30^{\circ}=20\times\frac{\sqrt3}{2}=17.32\,\text{m s}^{-1}$. Adding platform speed gives
Using \(\omega = \omega_0 + \alpha t\) with \(\omega_0 = 0\), \(\alpha = 2\) rad·s⁻² and \(t = 3\) s, we get \(\omega = 0 + 2\times3 = 6\) rad·s⁻¹....
Read Full Step-by-Step Solution →Vertical component: \(u_y = 30\sin30^{\circ}=15\) m/s, so time of flight \(T = 2u_y/g \approx 3.06\) s. Horizontal component relative to train: \(u_x ...
Read Full Step-by-Step Solution →The horizontal component of velocity remains constant, while the vertical component becomes zero at the maximum height....
Read Full Step-by-Step Solution →To find the horizontal distance covered, we need to find the horizontal component of the initial velocity and then use it to calculate the distance. T...
Read Full Step-by-Step Solution →The change in momentum is equal to the initial momentum, which is the product of the mass and velocity of the particle....
Read Full Step-by-Step Solution →At the highest point, the vertical component of velocity is 0. The initial vertical component of velocity is 50 sin(60°) = 25√3 m/s. Since the acceler...
Read Full Step-by-Step Solution →$H = \frac{u^2 \sin^2\theta}{2g} = \frac{400 \cdot (\sqrt{3}/2)^2}{20} = \frac{400 \cdot 0.75}{20} = 15\,\text{m}$...
Read Full Step-by-Step Solution →Horizontal component relative to ground: \(v_{x}=20\cos30^{\circ}+10 = 17.32+10 = 27.32\) m/s. Vertical component: \(v_{y}=20\sin30^{\circ}=10\) m/s. ...
Read Full Step-by-Step Solution →Relative velocity vector: $\vec{v}_{bg} = \vec{v}_{br} + \vec{v}_{rg}$. Magnitude = $\sqrt{3^2 + 4^2} = 5\,\text{km/h}$....
Read Full Step-by-Step Solution →Horizontal component relative to the platform: $u\cos30^{\circ}=20\times\frac{\sqrt3}{2}=17.32\,\text{m s}^{-1}$. Adding platform speed gives
Using \(\omega = \omega_0 + \alpha t\) with \(\omega_0 = 0\), \(\alpha = 2\) rad·s⁻² and \(t = 3\) s, we get \(\omega = 0 + 2\times3 = 6\) rad·s⁻¹....
Read Full Step-by-Step Solution →Vertical component: \(u_y = 30\sin30^{\circ}=15\) m/s, so time of flight \(T = 2u_y/g \approx 3.06\) s. Horizontal component relative to train: \(u_x ...
Read Full Step-by-Step Solution →The horizontal component of velocity remains constant, while the vertical component becomes zero at the maximum height....
Read Full Step-by-Step Solution →To find the horizontal distance covered, we need to find the horizontal component of the initial velocity and then use it to calculate the distance. T...
Read Full Step-by-Step Solution →The change in momentum is equal to the initial momentum, which is the product of the mass and velocity of the particle....
Read Full Step-by-Step Solution →At the highest point, the vertical component of velocity is 0. The initial vertical component of velocity is 50 sin(60°) = 25√3 m/s. Since the acceler...
Read Full Step-by-Step Solution →$H = \frac{u^2 \sin^2\theta}{2g} = \frac{400 \cdot (\sqrt{3}/2)^2}{20} = \frac{400 \cdot 0.75}{20} = 15\,\text{m}$...
Read Full Step-by-Step Solution →Horizontal component relative to ground: \(v_{x}=20\cos30^{\circ}+10 = 17.32+10 = 27.32\) m/s. Vertical component: \(v_{y}=20\sin30^{\circ}=10\) m/s. ...
Read Full Step-by-Step Solution →