Answers - Question 4
Part 1
Equipotentials are shown in a uniform electric field separated by 1 cm.
a Write down the field strength and the direction of the field.
The field strength is 2 Volts per cm ... or 200 Volts per meter. The field is perpendicular to the equipotentials to the left.
b Explain how the definition of field strength used above is extended when the field is not uniform.
The field strength is the negative gradient of the scalar potential . The - DV/ Dx. found above becomes the derivative -dV/dx.
b A spring is extended from zero extension and zero force to an extension x. The spring constant is k. The potential energy stored in the spring is 1/2 kx2.
Find the force required to extend the spring as a function of x using the principle above.
The force is the negative gradient of the scalar potential
= - d/dx (1/2 kx2) = - kx ... (Newtons) c A second spring is extended from zero with a force fo to an extension x. The spring constant is k. The potential energy stored in the spring is fox + 1/2 kx2.
Find the force required to extend the spring as a function of x using the principle above.
The force is the negative gradient of the scalar potential
= - d/dx (fox + 1/2 kx2) = - fo - kx ... (Newtons) Part 2
Force was defined by Newton as the rate of change of momentum.
a Explain how this concept applies when finding the maximum force on an object in a collision.
When objects collide the maximum force applied is the maximum rate of change of momentum. If the objects are hard the collision will last for a very short time and the rate of change of momentum will be very large.b Explain why the chances of survival are improved if a person falls into deep snow rather than into water.
If a person falls into deep snow the collision process is extended over a longer time than if they hit a hard water surface and the rate of change of momentum is lowered accordingly. The forces applied to the slowing body are therefore much less and cause less damage.
c A door is bombarded at right angles by a hail of 20 gram bullets traveling at 250 m/s. The bulets are embedded in the door. Find the average force on the door if 50 bullets per second hit the door.
The force on the door will depend on the rate of change of momentum. If the bullets are embedded in the door (stopped) the momentum charge is given by the momentum of one bullet times the number arriving per second.
= 0.02 x 250 x 50 = 250 Newtons at right angles to the door. d The door is now partly opened so that the door is at an angle of 45° to the bullets. What is now the average force on the door?
If the door is partly open the rate of change of momentum is the same (assuming the bullets are embedded in the door) and the force applied to the door will be the same in the direction of travel of the incoming bullets.
