1
JEE Main 2020 (Online) 9th January Evening Slot
+4
-1
A particle of mass m is projected with a speed u from the ground at an angle $$\theta = {\pi \over 3}$$ w.r.t. horizontal (x-axis). When it has reached its maximum height, it collides completely inelastically with another particle of the same mass and velocity $$u\widehat i$$ . The horizontal distance covered by the combined mass before reaching the ground is:
A
$$2\sqrt 2 {{{u^2}} \over g}$$
B
$${{3\sqrt 3 } \over 8}{{{u^2}} \over g}$$
C
$${{3\sqrt 2 } \over 4}{{{u^2}} \over g}$$
D
$${5 \over 8}{{{u^2}} \over g}$$
2
JEE Main 2020 (Online) 9th January Morning Slot
+4
-1
Two particles of equal mass m have respective
initial velocities $$u\widehat i$$ and $$u\left( {{{\widehat i + \widehat j} \over 2}} \right)$$.
They collide completely inelastically. The energy lost in the process is :
A
$${1 \over 3}m{u^2}$$
B
$${1 \over 8}m{u^2}$$
C
$${3 \over 4}m{u^2}$$
D
$$\sqrt {{2 \over 3}} m{u^2}$$
3
JEE Main 2020 (Online) 8th January Evening Slot
+4
-1
As shown in figure, when a spherical cavity (centered at O) of radius 1 is cut out of a uniform sphere of radius R (centered at C), the centre of mass of remaining (shaded) part of sphere is at G, i.e, on the surface of the cavity. R can be detemined by the equation :
A
(R2 + R – 1) (2 – R) = 1
B
(R2 – R – 1) (2 – R) = 1
C
(R2 – R + 1) (2 – R) = 1
D
(R2 + R + 1) (2 – R) = 1
4
JEE Main 2020 (Online) 8th January Evening Slot
+4
-1
A particle of mass m is dropped from a height h above the ground. At the same time another particle of the same mass is thrown vertically upwards from the ground with a speed of $$\sqrt {2gh}$$. If they collide head-on completely inelastically, the time taken for the combined mass to reach the ground, in units of $$\sqrt {{h \over g}}$$ is :
A
$$\sqrt {{1 \over 2}}$$
B
$${1 \over 2}$$
C
$$\sqrt {{3 \over 2}}$$
D
$$\sqrt {{3 \over 4}}$$
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