Two identical particles move towards each other with velocity 2v and vrespectively. The velocity of centre of mass is [AIEEE 2002]
Let say that direction of any one particle to be positive, let say right. Thus the equation of velocity for the center of mass = m x 2v + m x (-v) / m + m
= mv / 2m = v/2
Consider the following two statements :
A. Linear momentum of a system of particles is zero.
B. kinetic energy of a system of particles is zero.Then
We know momentum of a system = mass * change in velocity
And Kinetic Energy = 1/2 * mass * (velocity)²
If change in velocity is zero that doesn't mean that the velocity of the system is zero. If the velocity of a system is zero then there is no change in velocity.
A body A of mass M while falling vertically downwards under gravity breaks into two parts; a body B of mass M and, a body C of mass M. The centre of mass of bodies B and C taken together shifts compared to that of body A towards
Since there is no different external force applied to the individual particles, the centre of mass remains same and does not shift.
A mass m moves with a velocity v and collides inelastically with another identical mass. After collision the 1st mass moves with velocity in a direction perpendicular to the initial direction of motion. Find the speed of the second mass after collision.
Consider a two particle system with particles having masses m1 and m2. If the first particle is pushed towards the centre of mass through a distance d, by what distance should the second particle be moved, so as to keep the centre of mass at the same position?
Xcm = (m1x1 + m2x2)/(m1 + m2)
Xcm = m2L/(m1 + m2)
Xcm = (m1d + m2(L-x))/(m1 + m2)
Since Xcm is same so,
⇒ (m1d + m2(L-x))/(m1 + m2) = m2L/(m1 + m2)
⇒ m1d + m2L - m2x = m2L
⇒ m1d = m2x
⇒ x = m1d/m2
A bomb of mass 16 kg at rest explodes into two pieces of masses 4 kg and 12 kg. The velocity of the 12 kg mass is 4 ms-1. The kinetic energy of the other mass is
Applying Conservation of Linear Momentum
16 x 0 = 12 x 4 + 4 x v => v = -12 m/s
So, KE = ½ m v2 = ½ x 4 x 144 = 288J
A player caught a cricket ball of mass 150 g moving at a rate of 20 m/s. If the catching process is completed in 0.1 s, the force of the blow exerted by the ball on the hand of the player is equal to
F = ma = m(v/t)
F = 150 × 20 / 0.1 × 1000
F = 30N
A circular disc of radius R is removed from a bigger circular disc of radius 2 R, such that the circumference of the discs coincide. The centre of mass of the new disc is from the centre of the bigger disc. The value of a is
A thin rod of length L is lying along the x-axis with its ends at x = 0 and x = L. Its linear density (mass/length) varies with x as where n can be zero or any positive number. If the position of the centre of mass of the rod is plotted against n, which of the following graphs best approximates the dependence of on n?
A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is
Statement I : Two particles moving in the same direction do not lose all their energy in a completely inelastic collision.
Statement II : Principle of conservation of momentum holds true for all kinds of collisions.
The strong interaction force between two bodies in which momentum is conserved is called collision. In any type of collision momentum is always conserved kinetic energy may or may not be conserved.
Statement-I : A point particle of mass m moving with speed u collides with stationary point particle of mass M. If the maximum energy loss possible is given as then f = .
Statement-II : Maximum energy loss occurs when the particles get stuck together as a result of the collision. [JEE MAIN 2013]