Advertisements
Advertisements
प्रश्न
Two friends A and B (each weighing 40 kg) are sitting on a frictionless platform some distance d apart. A rolls a ball of mass 4 kg on the platform towards B which B catches. Then B rolls the ball towards A and A catches it. The ball keeps on moving back and forth between A and B. The ball has a fixed speed of 5 m/s on the platform. (a) Find the speed of A after he catches the ball for the first time. (c) Find the speeds of A and Bafter the all has made 5 round trips and is held by A. (d) How many times can A roll the ball? (e) Where is the centre of mass of the system "A + B + ball" at the end of the nth trip?
Advertisements
उत्तर
It is given that:
Weight of A = Weight of B = 40 kg
Velocity of ball = 5 m/s
(a) Case-1: Total momentum of the man A and ball remains constant.
∴ 0 = 4 × 5 − 40 × v
⇒ v = 0.5 m/s, towards left
(b) Case-2: When B catches the ball, the momentum between B and the ball remains constant.
⇒ 4 × 5 = 44 v
⇒ \[v = \left( \frac{20}{44} \right)\text{ m/s}\]
Case-3: When B throws the ball,
On applying the law of conservation of linear momentum, we get:
\[\Rightarrow 44 \times \left( \frac{20}{44} \right) = - 4 \times 5 + 40 \times v\]
\[ \Rightarrow v = 1 \text{ m/s , towards right)}\]
Case-4: When A catches the ball,
Applying the law of conservation of liner momentum, we get:
\[- 4 \times 5 + ( - 0 . 5) \times 40 = 44 v\]
\[ \Rightarrow v = \frac{40}{44} = \frac{10}{11}\text{ m/s, towards left}\]
(c) Case-5: When A throws the ball,
Applying the law of conservation of linear momentum, we get:
\[44 \times \left( \frac{10}{11} \right) = 4 \times 5 + 40 \times v\]
\[ \Rightarrow v = \frac{60}{40} = \frac{3}{2} \text{ m/s towards left)}\]
Case-6: When B receives the ball,
Applying the law of conservation of linear momentum, we get:
\[40 \times 1 + 4 \times 5 = 44 \times v\]
\[ \Rightarrow v = \frac{60}{44} \text{ m/s, towards right }\]
Case-7: When B throws the ball,
On applying the law of conservation of linear momentum, we get:
\[\Rightarrow v = 44 \times \left( \frac{60}{44} \right) \text{ m/s, towards right }\]
Case-8: When A catches the ball,
On applying the law of conservation of linear momentum, we get:
\[- 4 \times 5 + 40\left( \frac{3}{2} \right) = - 44v\]
\[ \Rightarrow V = - \frac{40}{44} = $\left( \frac{10}{11} \right) \text{ m/s,towards left } \]
Similarly, after 5 round trips,
The velocity of A will be \[\left( \frac{50}{11} \right)\] m/s and the velocity of B will be 5 m/s.
(d) As after 6 round trips, the velocity of A becomes \[\frac{60}{11}\] > 5 m/s, it cannot catch the ball. Thus, A can only roll the ball six times.
(e) Let the ball and the body A be at origin, in the initial position.
\[\therefore X_c = \frac{40 \times 0 + 4 \times 0 + 40 \times d}{40 + 40 + 4}\]
\[ = \frac{10}{21}d\]
APPEARS IN
संबंधित प्रश्न
If the total mechanical energy of a particle is zero, is its linear momentum necessarily zero? Is it necessarily nonzero?
Use the definition of linear momentum from the previous question. Can we state the principle of conservation of linear momentum for a single particle?
When a nucleus at rest emits a beta particle, it is found that the velocities of the recoiling nucleus and the beta particle are not along the same straight line. How can this be possible in view of the principle of conservation of momentum?
In one-dimensional elastic collision of equal masses, the velocities are interchanged. Can velocities in a one-dimensional collision be interchanged if the masses are not equal?
Consider the following two statements:
(A) The linear momentum of a particle is independent of the frame of reference.
(B) The kinetic energy of a particle is independent of the frame of reference.
A bullet hits a block kept at rest on a smooth horizontal surface and gets embedded into it. Which of the following does not change?
A nucleus moving with a velocity \[\vec{v}\] emits an α-particle. Let the velocities of the α-particle and the remaining nucleus be v1 and v2 and their masses be m1 and m2.
A shell is fired from a cannon with a velocity V at an angle θ with the horizontal direction. At the highest point in its path, it explodes into two pieces of equal masses. One of the pieces retraces its path to the cannon. The speed of the other piece immediately after the explosion is
A ball hits a floor and rebounds after an inelastic collision. In this case
(a) the momentum of the ball just after the collision is same as that just before the collision
(b) the mechanical energy of the ball remains the same during the collision
(c) the total momentum of the ball and the earth is conserved
(d) the total energy of the ball and the earth remains the same
A neutron initially at rest, decays into a proton, an electron, and an antineutrino. The ejected electron has a momentum of 1.4 × 10−26 kg-m/s and the antineutrino 6.4 × 10−27kg-m/s.
Find the recoil speed of the proton
(a) if the electron and the antineutrino are ejected along the same direction and
(b) if they are ejected along perpendicular directions. Mass of the proton = 1.67 × 10−27 kg.
A man of mass M having a bag of mass m slips from the roof of a tall building of height H and starts falling vertically in the following figure. When at a height h from the ground, the notices that the ground below him is pretty hard, but there is a pond at a horizontal distance x from the line of fall. In order to save himself he throws the bag horizontally (with respect to himself) in the direction opposite to the pond. Calculate the minimum horizontal velocity imparted to the bag so that the man lands in the water. If the man just succeeds to avoid the hard ground, where will the bag land?
A gun is mounted on a railroad car. The mass of the car, the gun, the shells and the operator is 50 m where m is the mass of one shell. If the velocity of the shell with respect to the gun (in its state before firing) is 200 m/s, what is the recoil speed of the car after the second shot? Neglect friction.
Consider a head-on collision between two particles of masses m1 and m2. The initial speeds of the particles are u1 and u2 in the same direction. the collision starts at t = 0 and the particles interact for a time interval ∆t. During the collision, the speed of the first particle varies as \[v(t) = u_1 + \frac{t}{∆ t}( v_1 - u_1 )\]
Find the speed of the second particle as a function of time during the collision.
A bullet of mass 20 g travelling horizontally with a speed of 500 m/s passes through a wooden block of mass 10.0 kg initially at rest on a level surface. The bullet emerges with a speed of 100 m/s and the block slides 20 cm on the surface before coming to rest. Find the friction coefficient between the block and the surface (See figure).
A block of mass 200 g is suspended through a vertical spring. The spring is stretched by 1.0 cm when the block is in equilibrium. A particle of mass 120 g is dropped on the block from a height of 45 cm. The particle sticks to the block after the impact. Find the maximum extension of the spring. Take g = 10 m/s2.
A bullet of mass 25 g is fired horizontally into a ballistic pendulum of mass 5.0 kg and gets embedded in it. If the centre of the pendulum rises by a distance of 10 cm, find the speed of the bullet.
A small block of superdense material has a mass of 3 × 1024kg. It is situated at a height h (much smaller than the earth's radius) from where it falls on the earth's surface. Find its speed when its height from the earth's surface has reduce to to h/2. The mass of the earth is 6 × 1024kg.
The track shown is figure is frictionless. The block B of mass 2m is lying at rest and the block A or mass m is pushed along the track with some speed. The collision between Aand B is perfectly elastic. With what velocity should the block A be started to get the sleeping man awakened?
