1
JEE Main 2018 (Offline)
MCQ (Single Correct Answer)
+4
-1
Change Language
In a collinear collision, a particle with an initial speed v0 strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is :
A
$${{{v_0}} \over {\sqrt 2 }}$$
B
$${{v_0}} \over 4$$
C
$$\sqrt 2 {v_0}$$
D
$${{v_0}} \over 2$$
2
JEE Main 2018 (Online) 15th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A proton of mass m collides elastically with a particle of unknown mass at rest. After the collision, the proton and the unknown particle are seen moving at an angle of 90o with respect to each other. The mass of unknown particle is :
A
$${m \over 2}$$
B
m
C
$${m \over {\sqrt 3 }}$$
D
2 m
3
JEE Main 2017 (Online) 9th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Two particles A and B of equal mass M are moving with the same speed $$\upsilon $$ as shown in the figure. They collide completely inelastically and move as a single particle C. The angle $$\theta $$ that the path of C makes with the X-axis is given by :

JEE Main 2017 (Online) 9th April Morning Slot Physics - Center of Mass and Collision Question 103 English
A
tan $$\theta $$ = $${{\sqrt 3 + \sqrt 2 } \over {1 - \sqrt 2 }}$$
B
tan $$\theta $$ = $${{\sqrt 3 - \sqrt 2 } \over {1 - \sqrt 2 }}$$
C
tan $$\theta $$ = $${{1 - \sqrt 2 } \over {\sqrt 2 \left( {1 + \sqrt 3 } \right)}}$$
D
tan $$\theta $$ = $${{1 - \sqrt 3 } \over {1 + \sqrt 2 }}$$
4
JEE Main 2016 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
In the figure shown ABC is a uniform wire. If centre of mass of wire lies vertically below point A, then $${{BC} \over {AB}}$$ is close to :

JEE Main 2016 (Online) 10th April Morning Slot Physics - Center of Mass and Collision Question 101 English
A
1.85
B
1.37
C
1.5
D
3
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