Home
Class 11
PHYSICS
Two particles of mases m(1) and m(2) in ...

Two particles of mases `m_(1)` and `m_(2)` in projectile motion have velocitie `vecv_(1)` and `vecv_(2)`, respectively at time `t=0`. they collide at time `t_(0)`. Their velocities become `vecv_(1)^(')` and `vecv_(2)^(')` at time `2t_(0)` while still moving in air. The value of
`|(m_(1)vecv_(1)^(')+m_(2)vecv_(2)^('))|-|(m_(1)vecv_(1)+m_(2)vecv_(2))|` is

A

zero

B

`(m_(1) + m_(2)) g t_(0)`

C

`2(m_(1) + m_(2))g t_(0)`

D

`(1)/(2)(m_(1) + m_(2)) g t_(0)`

Text Solution

Verified by Experts

The correct Answer is:
C
Promotional Banner

Similar Questions

Explore conceptually related problems

Two particles of masses m_(1) and m_(2) in projectile motion have velocities vec(v)_(1) and vec(v)_(2) , respectively , at time t = 0 . They collide at time t_(0) . Their velocities become vec(v')_(1) and vec(v')_(2) at time 2 t_(0) while still moving in air. The value of |(m_(1) vec(v')_(1) + m_(2) vec(v')_(2)) - (m_(1) vec(v)_(1) + m_(2) vec(v)_(2))|

Two particles of same mass 'm' moving with velocities vecv_1= vhati, and vecv_2 = (v/2)hati + (v/2)hatj collide in-elastically. Find the loss in kinetic energy

Two particles of same mass 'm' moving with velocities vecv_1= vhati, and vecv_2 = (v/2)hati + (v/2)hatj collide in-elastically. Find the loss in kinetic energy

Assertion : An object has given two velocities vecv_(1) and vecv_(2) has a resultant velocity vecv = vecv_(1) + vecv_(2) . Reason : vecv_(1) and vecv_(2) should be velocities with reference to some common reference frame.

Two satellites A and B of masses m_(1) and m_(2)(m_(1)=2m_(2)) are moving in circular orbits of radii r_(1) and r_(2)(r_(1)=4r_(2)) , respectively, around the earth. If their periods are T_(A) and T_(B) , then the ratio T_(A)//T_(B) is

The corresponding quantities in rotational motion related to m, vecF, vecp and vecv in linear motion are respectively

The velociies of A and B are vecv_(A)= 2hati + 4hatj and vecv_(B) = 3hati - 7hatj , velocity of B as observed by A is

Two particles having position verctors vecr_(1)=(3hati+5hatj) metres and vecr_(2)=(-5hati-3hatj) metres are moving with velocities vecv_(1)=(4hati+3hatj)m//s and vecv_(2)=(alphahati+7hatj)m//s . If they collide after 2 seconds, the value of alpha is

A particle of mass 1 g moving with a velocity vecv_1=3hati-2hatj m s^(-1) experiences a perfectly in elastic collision with another particle of mass 2 g and velocity vecv_2=4hatj-6 hatk m s^(-1) . The velocity of the particle is

The ends A and B of a eod of length l have velocities of magnitudes |vecv_(A)|=v and |vecv_(B)|=2v respectively. If the inclination of vecv_(A) relationn to the rod is alpha find the a. Inclination beta of vecc_(B) relative of the rod. b. angular velocity of the rod.