A particle of mass m moving with velocity vec | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For an elastic collision between two particles of equal mass where one is initially at rest,the conservation of linear momentum is given by: $m\ve

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(C) For an elastic collision between two particles of equal mass where one is initially at rest,the conservation of linear momentum is given by: $m\vec{v_1} = m\vec{v_1}' + m\vec{v_2}'$,which simplifies to $\vec{v_1} = \vec{v_1}' + \vec{v_2}'$.Taking the dot product of both sides with themselves: $\vec{v_1} \cdot \vec{v_1} = (\vec{v_1}' + \vec{v_2}') \cdot (\vec{v_1}' + \vec{v_2}')$.This results in: $v_1^2 = v_1'^2 + v_2'^2 + 2(\vec{v_1}' \cdot \vec{v_2}')$ ... $(1)$.From the law of conservation of kinetic energy: $\frac{1}{2}mv_1^2 = \frac{1}{2}mv_1'^2 + \frac{1}{2}mv_2'^2$,which simplifies to $v_1^2 = v_1'^2 + v_2'^2$ ... $(2)$.Substituting equation $(2)$ into equation $(1)$: $v_1'^2 + v_2'^2 = v_1'^2 + v_2'^2 + 2(\vec{v_1}' \cdot \vec{v_2}')$.This implies $2(\vec{v_1}' \cdot \vec{v_2}') = 0$,so $\vec{v_1}' \cdot \vec{v_2}' = 0$.Since the dot product is $v_1' v_2' \cos 	heta = 0$,and the velocities are non-zero,we have $\cos 	heta = 0$,which means $	heta = 90^\circ$.

$A$ particle of mass $m$ moving with velocity $\vec{v_1}$ undergoes a two-dimensional elastic collision with another particle of mass $m$ at rest. If the particles move with velocities $\vec{v_1}'$ and $\vec{v_2}'$ after the collision,what is the angle between them in degrees?

Similar Questions

$A$ neutron moving with velocity $u$ collides elastically with an atom of mass number $A$. If the collision is head-on and the initial kinetic energy of the neutron is $E$, then the final kinetic energy of the neutron after the collision is:

$A$ body of mass $m$ moving with velocity $v$ makes a head-on elastic collision with another body of mass $2m$ which is initially at rest. What is the loss in kinetic energy of the colliding body?

$A$ ball of mass $m$ moving with a speed $u$ undergoes a head-on elastic collision with a ball of mass $nm$ initially at rest. The fraction of the incident energy transferred to the heavier ball is

$A$ ball of mass $10\, kg$ is moving with a velocity of $10\, m/s$. It strikes another ball of mass $5\, kg$ which is moving in the same direction with a velocity of $4\, m/s$. If the collision is elastic,their velocities after the collision will be,respectively

$A$ block $A$ of mass $M$ moving with speed $u$ collides elastically with block $B$ of mass $m$ which is connected to block $C$ of mass $m$ with a spring. When the compression in the spring is maximum,the velocity of block $C$ with respect to block $A$ is (neglect friction):

Vedclass Test Series

Exam Paper Generator

Online Exam Module