Two particles of masses m_{1} and m_{2} in projectile motion have velocities respectively at time t = 0. They collide at time t_{o}. Their velocities become at time 2t_{o} while still moving in air. The value of
If we consider the two particles as a system then the external force acting on the system is the gravitational pull (m_{1} + m_{2}) g.
Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the centre of mass is
Just after collision
Note : Spring force is an internal force, it cannot change the linear momentum of the (two mass + spring) system.
Therefore v_{c} remains the same.
A ball of mass 0.2 kg rests on a vertical post of height 5 m. A bullet of mass 0.01 kg, traveling with a velocity V m/s in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 m and the bullet at a distance of 100 m from the foot of the post. The velocity V of the bullet is
For vertical motion of bullet or ball
u = 0, s = 5m, t = ? , a = 10m/s^{2}
For horizontal motion of ball
For horizontal motion of bullet
Applying conservation of linear momentum during collision, w e get
A particle of mass m is projected from the ground with an initial speed u_{0} at an angle α with the horizontal. At the highest point of its trajectory, it makes a completely inelastic collision with another identical particle, which was thrown vertically upward from the ground with the same initial speed u_{0}. The angle that the composite system makes with the horizontal immediately after the collision is
Activity B to M for particle thrown upwards
Applying conservation of linear momentum in Ydirection
2mv sinθ = mv1 = mu_{0}cosα ...(ii) [from (i)]
Applying conservation of linear momentum in Xdirection
2mv cosθ = mu_{0} cosα ...(iii)
on dividing (ii) and (iii) we get
A ball hits the floor and rebounds after an inelastic collision. In this case
(a) is wrong because the momentum of ball changes in magnitude as well as direction.
(b) is wrong because on collision, some mechanical energy is converted into heat, sound energy.
(c) is correct because for earth + ball system the impact force is an internal force.
(d) is correct.
A shell is fired from a cannon with a velocity v (m/sec.) at an angle θ with the horizontal direction. At the highest point in its path it explodes into two pieces of equal mass. One of the pieces retraces its path to the cannon and the speed (in m/sec.) of the other piece immediately after the explosion is
As one piece retraces its path, the speed of this piece just after explosion should be v cos θ
(At highest point just after explosion) NOTE THIS STEP
Applying conservation of linear momentum at the highest point;
Two blocks A and B, each of mass m, are connected by a massless spring of natural length L and spring constant K.
The blocks are initially resting on a smooth horizontal floor with the spring at its natural length, as shown in fig.. A third identical block C, also of mass m, moves on the floor with a speed v along the line joining A and B, and collides elastically with A. Then
In situation 1, mass C is moving towards right with velocity v. A and B are at rest.
In situation 2, which is just after the collision of C with A, C stop and A acquires a velocity v. [headon elastic collision between identical masses]
When A starts moving towards right, the spring suffer a compression due to which B also starts moving towards right. The compression of the spring continues till there is relative velocity between A and B. When this relative velocity becomes zero, both A and B move with the same velocity v' and the spring is in a state of maximum compression.
Applying momentum conservation in situation 1 and 3,
∴ K.E. of the system in situation 3 is
This is the kinetic energy possessed by A – B system (since, C is at rest).
Let x be the maximum compression of the spring.
Applying energy conservation
The balls, having linear momenta undergo a collision in free space. There is no external force acting on the balls. Let be their final momenta. The following option (s) is (are) NOT ALLOWED for any nonzero value of p, a_{1}, a_{2}, b_{1}, b_{2}, c_{1} and c_{2}.
KEY CONCEPT Use law of conservation of linear momentum.
The initial linear momentum of the system is Therefore the final linear momentum should also be zero.
Option a :
Final momentum.
It is given that a_{1}, b_{1}, c_{1}, a_{2}, b_{2} and c_{2} have nonzero values. If a_{1} = x and a_{2} = – x. Also if b_{1} = y and b_{2} = – y then the components become zero. But the third term having component is nonzero. This gives a definite final momentum to the system which violates conservation of linear momentum, so this is an incorrect option.
Option d:
because b_{1} ≠ 0
Following the same reasoning as above this option is also ruled ou
A point mass of 1 kg collides elastically with a stationary point mass of 5 kg. After their collision, the 1 kg mass reverses its direction and moves with a speed of 2 ms^{–1}. Which of the following statement(s) is (are) correct for the system of these two masses?
According to law of conservation of linear momentum 1 × u_{1} + 5 × 0 = 1 (–2) + 5 (v_{2})
⇒ u_{1} = – 2 + 5 v_{2} ...(i)
The coefficient of restituition
On solving (i) & (ii) we get desired results.
A particle of mass m is attached to one end of a massless spring of force constant k, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time t = 0 with an initial velocity u_{0}. When the speed of the particle is 0.5 u_{0}, it collides elastically with a rigid wall. After this collision
The particle collides elastically with rigid wall. Here
i.e. the particle rebounds with the same speed. Therefore the particle will return to its equilibrium position with speed u_{0}. option (a) is correct.
The velocity of the particle becomes 0.5u_{0} after time t.
Then using the equation V = V_{max} cos wt we get 0.5u_{0} = u_{0} cos wt
The time period
The time taken by the particle to pass through the equilibrium for the first time Therefore option (b) is incorrect
The time taken for the maximum compression
Therefore option c is incorrect.
The time taken for particle to pass through the equilibrium position second time
option (d) is correct.
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