1
JEE Main 2019 (Online) 12th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
An alpha-particle of mass m suffers 1-dimensional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing, 64% of its initial kinetic energy. The mass of the nucleus is
A
2m
B
4m
C
1.5m
D
3.5m
2
JEE Main 2019 (Online) 12th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The position vector of the centre of mass $$\overrightarrow r {\,_{cm}}\,$$ of an asymmetric uniform bar of negligible area of crosssection as shown in figure is :

JEE Main 2019 (Online) 12th January Morning Slot Physics - Center of Mass and Collision Question 90 English
A
$${\overrightarrow r _{cm}}\, = {{11} \over 8}L\,\,\widehat x + {8 \over 8}L\widehat y$$
B
$${\overrightarrow r _{cm}}\, = {5 \over 8}L\,\,\widehat x + {{13} \over 8}L\widehat y$$
C
$${\overrightarrow r _{cm}}\, = {{13} \over 8}L\,\,\widehat x + {5 \over 8}L\widehat y$$
D
$${\overrightarrow r _{cm}}\, = {3 \over 8}L\,\,\widehat x + {{11} \over 8}L\widehat y$$
3
JEE Main 2019 (Online) 12th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A simple pendulum, made of a string of length $$\ell $$ and a bob of mass m, is released from a small angle $${{\theta _0}}$$. It strikes a block of mass M, kept on a horizontal surface at its lowest point of oscillations, elastically. It bounces back and goes up to an angle $${{\theta _1}}$$. Then M is given by :
A
$${m \over 2}\left( {{{{\theta _0} + {\theta _1}} \over {{\theta _0} - {\theta _1}}}} \right)$$
B
$${m \over 2}\left( {{{{\theta _0} - {\theta _1}} \over {{\theta _0} + {\theta _1}}}} \right)$$
C
$$m\left( {{{{\theta _0} + {\theta _1}} \over {{\theta _0} - {\theta _1}}}} \right)$$
D
$$m\left( {{{{\theta _0} - {\theta _1}} \over {{\theta _0} + {\theta _1}}}} \right)$$
4
JEE Main 2019 (Online) 11th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A particle of mass m is moving in a straight line with momentum p. Starting at time t = 0, a force F = kt acts in the same direction on the moving particle during time interval T so that its momentum changes from p to 3p. Here k is a constant. The value of T is :
A
$$2\sqrt {{k \over p}} $$
B
$$2\sqrt {{p \over k}} $$
C
$$\sqrt {{{2p} \over 2}} $$
D
$$\sqrt {{{2k} \over p}} $$
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