Advertisements
Advertisements
प्रश्न
A bullet of mass m fired at 30° to the horizontal leaves the barrel of the gun with a velocity v. The bullet hits a soft target at a height h above the ground while it is moving downward and emerges out with half the kinetic energy it had before hitting the target.
Which of the following statements are correct in respect of bullet after it emerges out of the target?
- The velocity of the bullet will be reduced to half its initial value.
- The velocity of the bullet will be more than half of its earlier velocity.
- The bullet will continue to move along the same parabolic path.
- The bullet will move in a different parabolic path.
- The bullet will fall vertically downward after hitting the target.
- The internal energy of the particles of the target will increase.
Advertisements
उत्तर
b, d and f
Explanation:
Consider the adjacent diagram for the given situation in the question.
b. Conserving energy between "O" and "A"
`U_i + K_i = U_f + K_f`
⇒ `0 + 1/2 mv^2 = mgh + 1/2 mv^'`
⇒ `(v^')^2/2 = v^2/2 = - gh`
⇒ `(v^')^2 = v^2 - 2 gh`
⇒ `v^' = sqrt(v^2 - 2 gh)` ......(i)
Where v' is the speed of the bullet just before hitting the target. Let speed after emerging from the target is v" then,
By question, = `1/2 (mv^")^2 = 1/2[1/2 m(v^')^2]`
= `1/2 m(v^")^2`
= `1/4 m(v^')^2`
= `1/4 m[v^2 - 2 gh]`
⇒ `(v^")^2 = (v^2 - 2 gh)/2 = v^2/2 - gh`
⇒ `v^" = sqrt(v^2/2 - gh)` .....(ii)
From equations (i) and (ii)
`v^'/v^" = (sqrt(v^2 - 2 gh))/(sqrt(v^2 - 2 gh)/(sqrt(2))) = sqrt(2)`
⇒ `v^" = v^'/sqrt(2) = v^2 (v^'/2)`
⇒ `v^"/(v^'/2) = sqrt(2)` = 1.414 > 1
⇒ `v^" > v^'/2`
Hence, after emerging from the target velocity of the bullet (v") is more than half of its earlier velocity v’ (velocity before emerging into the target).
d. As the velocity of the bullet changes to v’ which is less than v1 hence, the path, followed will change and the bullet reaches point B instead of A', as shown in the figure.
f. As the bullet is passing through the target the loss in energy of the bullet is transferred to particles of the target. Therefore, their internal energy increases.
APPEARS IN
संबंधित प्रश्न
In figure (i) the man walks 2 m carrying a mass of 15 kg on his hands. In Figure (ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater?
Is work-energy theorem valid in non-inertial frames?
Consider the situation of the previous question from a frame moving with a speed v0 parallel to the initial velocity of the block. (a) What are the initial and final kinetic energies? (b) What is the work done by the kinetic friction?
A water pump lifts water from 10 m below the ground. Water is pumped at a rate of 30 kg/minute with negligible velocity. Calculate the minimum horsepower that the engine should have to do this.
A block of mass 30 kg is being brought down by a chain. If the block acquires a speed of 40 cm/s in dropping down 2 m, find the work done by the chain during the process.
A small block of mass 200 g is kept at the top of a frictionless incline which is 10 m long and 3⋅2 m high. How much work was required (a) to lift the block from the ground and put it an the top, (b) to slide the block up the incline? What will be the speed of the block when it reaches the ground if (c) it falls off the incline and drops vertically to the ground (d) it slides down the incline? Take g = 10 m/s2.
A block of mass 5 kg is suspended from the end of a vertical spring which is stretched by 10 cm under the load of the block. The block is given a sharp impulse from below, so that it acquires an upward speed of 2 m/s. How high will it rise? Take g = 10 m/s2.
The bob of a pendulum at rest is given a sharp hit to impart a horizontal velocity \[\sqrt{10 \text{ gl }}\], where l is the length of the pendulum. Find the tension in the string when (a) the string is horizontal, (b) the bob is at its highest point and (c) the string makes an angle of 60° with the upward vertical.
Following figure following shows a smooth track, a part of which is a circle of radius R. A block of mass m is pushed against a spring of spring constant k fixed at the left end and is then released. Find the initial compression of the spring so that the block presses the track with a force mg when it reaches the point P, where the radius of the track is horizontal.
The bob of a stationary pendulum is given a sharp hit to impart it a horizontal speed of \[\sqrt{3 gl}\] . Find the angle rotated by the string before it becomes slack.
A heavy particle is suspended by a 1⋅5 m long string. It is given a horizontal velocity of \[\sqrt{57} \text{m/s}\] (a) Find the angle made by the string with the upward vertical when it becomes slack. (b) Find the speed of the particle at this instant. (c) Find the maximum height reached by the particle over the point of suspension. Take g = 10 m/s2.
A particle of mass m is kept on the top of a smooth sphere of radius R. It is given a sharp impulse which imparts it a horizontal speed ν. (a) Find the normal force between the sphere and the particle just after the impulse. (b) What should be the minimum value of ν for which the particle does not slip on the sphere? (c) Assuming the velocity ν to be half the minimum calculated in part, (b) find the angle made by the radius through the particle with the vertical when it leaves the sphere.
A chain of length l and mass m lies on the surface of a smooth sphere of radius R > l with one end tied to the top of the sphere. Find the gravitational potential energy of the chain with reference level at the centre of the sphere.
A chain of length l and mass m lies on the surface of a smooth sphere of radius R > l with one end tied to the top of the sphere. Find the tangential acceleration \[\frac{d\nu}{dt}\] of the chain when the chain starts sliding down.
A smooth sphere of radius R is made to translate in a straight line with a constant acceleration a. A particle kept on the top of the sphere is released at zero velocity with respect to the sphere. Find the speed of the particle with respect to the sphere as a function of the angle θ it slides.
A man, of mass m, standing at the bottom of the staircase, of height L climbs it and stands at its top.
- Work done by all forces on man is equal to the rise in potential energy mgL.
- Work done by all forces on man is zero.
- Work done by the gravitational force on man is mgL.
- The reaction force from a step does not do work because the point of application of the force does not move while the force exists.
Give example of a situation in which an applied force does not result in a change in kinetic energy.
Suppose the average mass of raindrops is 3.0 × 10–5 kg and their average terminal velocity 9 ms–1. Calculate the energy transferred by rain to each square metre of the surface at a place which receives 100 cm of rain in a year.
