If the collision of the ball with the building is elastic then find the angle at which the ball will rebound from the top of the building
Answers
In physics, the term conservation refers to something which doesn't change. This means that the variable in an equation which represents a conserved quantity is constant over time. It has the same value both before and after an event.
There are many conserved quantities in physics. They are often remarkably useful for making predictions in what would otherwise be very complicated situations. In mechanics, there are three fundamental quantities which are conserved. These are momentum, energy, and angular momentum. Conservation of momentum is mostly used for describing collisions between objects.
Just as with the other conservation principles, there is a catch: conservation of momentum applies only to an isolated system of objects. In this case an isolated system is one that is not acted on by force external to the system—i.e., there is no external impulse. What this means in the practical example of a collision between two objects is that we need to include both objects and anything else that applies a force to any of the objects for any length of time in the system.
If the subscripts ii and ff denote the initial and final momenta of objects in a system, then the principle of conservation of momentum says
=0.25 kg is swung and collides with a stationary golf ball of mass m_b=0.05~\mathrm{kg}m
b
=0.05 kg. High speed video shows that the club is traveling at v_c = 40~\mathrm{m/s}v
c
=40 m/s when it touches the ball. It remains in contact with the ball for t=0.5~\mathrm{ms}t=0.5 ms; after that, the ball is traveling at a speed of v_b=40~\mathrm{m/s}v
b
=40 m/s. How fast is the club traveling after it has hit the ball? [Show solution.]
p
\begin{aligned}p &= 0.25~\mathrm{kg}\cdot 40 ~ \mathrm{m/s} \\ &= 10~\mathrm{kg\cdot m/s}\end{aligned}
p–mbvbmc=vc=32 m/s
Exercise 2b: What is the average force on the club due to the golf ball in the previous problem? [Show solution.]
\begin{aligned} a &= \frac{\Delta v}{\Delta t} \\ \\\\ &=\dfrac{32\,\dfrac{\text m}{\text s} -40\,\dfrac{\text m}{\text s}}{0.5 \cdot 10^{-3} \,\text s}\\ \\\\ &= \frac{-8.0~\mathrm{m/s}}{0.5 \cdot 10^{-3} ~ \mathrm{s}} \\ \\\\ &= -16 \cdot 10^3 ~ \mathrm{m/s^2} \end{aligned}
Answer:
180
Explanation: