Show that `J = mDeltaV`, where J is the impulse acting on a body of mass ma nd `DeltaV` is the change in velocity.

A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s^-1 to 3.5 m s^-1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

Check whether equation F.S. = 1/2 mv^2 -1/2 m u^2 is dimensionally correct, where m is mass of the body, v its final velocity, u its initial velocity, F is force applied and S is the distance moved.

A body of mass 2kg moving with velocity 10 m/s collides another body of mass 500 g and moving with velocity 4 m/s. After collision if the 2nd body attains velocity of 8m/s, what will be the velocity of first body?

The kinetic energy of an object of mass, m moving with a elocity of 5 ms^-1 , is 25J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times?

If the mass of a body is halved and its velocity doubled, its momentum will not change. Explain.

Show thaqt the current I through an area of cross section is given by the scalaer product of two vectors I=jdelta S where j and deltas and vectors.

A body of mass 25 g is moving with a constant velocity of 5m/sec on a horizontal frictionless surface in vacuum. What is the force acting on the body?

A calorie is a unit of heat or energy and it equals about 4.2 J where 1J = 1 kg m^2s^(-2)c . Suppose we employ a system of units in which the unit of mass equals a kg, the unit of length equals R m, the unit of time y s? Show that a calorie has a magnitude 4.2 a^(-1) beta(-2) gamma2 in terms of the new units.

A force of 200 N is applied to a body and its velocity changes from 5m/s to 10m/s in a second. What is the mass of the body?

The potential energy function for a particle executing linear simple harmonic motion is given by V(x) =kx^2/2, where k is the force constant of the oscillator. For k = 0.5 N ^-1 , the graph of V(x) versus x is shown in Fig. 6.12. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m.,br>