Answer:
- Work done is maximum when the movement of object is in line and direction of force.
OR
- Work done is maximum, when displacement takes place along the direction of force.
- Work done is given by the equation
W = F.S
<em> W = F. S cos Θ</em>
<em>When cos Θ = 0° ; cos 0 = 1</em>
Answer:
The final kinetic energy of the Helium nucleus (alpha particle) after been scattered through an angle of 120° is
8.00 x 10-13J
Explanation:
In Rutherford Scattering experiment, the collision of the helium nucleus with the gold nucleus is an ELASTIC COLLISION. This means that the kinetic energy is conserved ( The same before and after the collision).
Thus, the final kinetic energy of the helium nucleus is the same as initial kinetic energy (8.00 x 10^-13Joules)
Although, the kinetic energy is converted to potential energy in Coulomb's law equation.
That is,
1/2(mv^2) = (K* q1q2)/r
Where m is the mass of helium nucleus, v is its colliding velocity, k is electrostatic constant, q1 is the charge on helium nucleus, q2 is the charge on gold nucleus, r is impact parameter
Maybe you can split up the questions. I will try to answer your first question.
1. In an elastic collision, momentum is conserved. The momentum before the collision is equal to the momentum after the collision. This is a consequence of Newton's 3rd law. (Action = Reaction)
2. Momentum: p = m₁v₁ + m₂v₂
m₁ mass of ball A
v₁ velocity of ball A
m₂ mass of ball B
v₂ velocity of ball B
Momentum before the collision:
p = 2*9 + 3*(-6) = 18 - 18 = 0
Momentum after the collision:
p = 2*(-9) + 3*6 = -18 + 18 = 0
3: mv + m(-v) = m(-v) + m(v)
the velocities would reverse.
4.This question is not factual since the energy of an elastic collision must also be conserved. The final velocities should be: v₁ = -1 m/s and v₂ = 5 m/s. That said assuming the given velocities were correct:
before collision
p = 10*3 + 5*(-3) = 30 - 15 = 15
after collision:
p = 10*(-2) + 5 * v₂ = 15
v₂ = 7
5.You figure out.
The correct graph is <u>D</u>.
The graph <em>A</em> is a straight line sloping downwards and it shows that the speed of the body is decreasing at a constant rate. Therefore, this s a graph of a body that is under a constant deceleration.
The graph B is a straight line which slopes upwards. Hence the graph shows that the speed of the body increases at a constant rate. Therefore, this is a graph of a body that is accelerating at a constant rate.
The graph C is curved line, which curves upwards. The slope of the curve increases with time. This is therefore, a graph of a body which is under increasing acceleration.
The graph D, however is a straight line parallel to the time axis. The speed of the body has the same value at all times. Therefore, Graph D is the graph which shows the motion of a body with constant speed.
