
A sphere collides with another sphere of identical mass. After collision, the two spheres move. The collision is inelastic. Then the angle between the directions of the two spheres is
A. Different from ${{90}^{\circ }}$
B. ${{90}^{\circ }}$
C. ${{0}^{\circ }}$
D. ${{45}^{\circ }}$
Answer
162k+ views
Hint: While the other body is at rest, the other body is in motion. As a result, the first body will have momentum, and the second body will have zero momentum since, as was already mentioned, it is at rest. We can now determine the answer using momentum and energy conservation.
Complete step by step solution:
An elastic collision is one in which the system does not experience a net loss of kinetic energy as a result of the collision. In an elastic collision, both momentum and kinetic energy are conserved.
\[{m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}\]
Where,
\[{m_1},{m_2} = \] Object’s mass
\[{u_1},{u_2} = \]Initial velocity
\[{v_1},{v_2} = \]Final velocity
In order for two bodies to collide exactly elastically, their paths must be at an angle of \[{90^ \circ }\]. When some of the kinetic energy of a colliding object or system is wasted, the collision is said to be inelastic.
\[{m_1}{u_1} + {m_2}{u_2} = \left( {{m_1} + {m_2}} \right){v_f}\]
Where,
\[{m_1},{m_2} = \] Object’s mass
\[{u_1},{u_2} = \]Initial velocity
\[{v_f} = \]Final velocity
As a result, the angle for an inelastic collision should be different from \[{90^ \circ }\]. A sphere with identical mass collides with each other. The two spheres move once they have collided. There is no elastic collision. If the collision is perfectly elastic, the angle between the directions of the two spheres will be \[90\] degrees.
Hence, option A is correct.
Note: We are aware that whenever the body moves, the body should also experience momentum, and vice versa. Mass and velocity are multiplied to create momentum. We now understand that the second sphere has no momentum prior to impact. Following the impact, the second will feel momentum as a result of the movement it receives. Therefore, we can simply answer this question using momentum conservation.
Complete step by step solution:
An elastic collision is one in which the system does not experience a net loss of kinetic energy as a result of the collision. In an elastic collision, both momentum and kinetic energy are conserved.
\[{m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}\]
Where,
\[{m_1},{m_2} = \] Object’s mass
\[{u_1},{u_2} = \]Initial velocity
\[{v_1},{v_2} = \]Final velocity
In order for two bodies to collide exactly elastically, their paths must be at an angle of \[{90^ \circ }\]. When some of the kinetic energy of a colliding object or system is wasted, the collision is said to be inelastic.
\[{m_1}{u_1} + {m_2}{u_2} = \left( {{m_1} + {m_2}} \right){v_f}\]
Where,
\[{m_1},{m_2} = \] Object’s mass
\[{u_1},{u_2} = \]Initial velocity
\[{v_f} = \]Final velocity
As a result, the angle for an inelastic collision should be different from \[{90^ \circ }\]. A sphere with identical mass collides with each other. The two spheres move once they have collided. There is no elastic collision. If the collision is perfectly elastic, the angle between the directions of the two spheres will be \[90\] degrees.
Hence, option A is correct.
Note: We are aware that whenever the body moves, the body should also experience momentum, and vice versa. Mass and velocity are multiplied to create momentum. We now understand that the second sphere has no momentum prior to impact. Following the impact, the second will feel momentum as a result of the movement it receives. Therefore, we can simply answer this question using momentum conservation.
Recently Updated Pages
A steel rail of length 5m and area of cross section class 11 physics JEE_Main

At which height is gravity zero class 11 physics JEE_Main

A nucleus of mass m + Delta m is at rest and decays class 11 physics JEE_MAIN

A wave is travelling along a string At an instant the class 11 physics JEE_Main

The length of a conductor is halved its conductivity class 11 physics JEE_Main

Two billiard balls of the same size and mass are in class 11 physics JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Class 11 JEE Main Physics Mock Test 2025

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

Other Pages
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
