Grandfather would be mad

The piece of gum possesses kinetic energy which it shares with the pendulum after impact. To find out the velocity of the system after the inelastic collision we will have to equate the momentum of the system before and after the collision,

$0.01\times10 + 0.09\times0 = 0.1 \times v$

$0.1 = 0.1v$

$v = 1 m/s$ .

Knowing the velocity of the system after the collision, we equate the kinetic energy of the system to the gravitational potential energy to find the height of the pendulum where all the kinetic energy of the system is converted to gravitational potential energy i.e. the highest point of the pendulum. i.e. $mgh = 0.5mv^2 \rightarrow gh=0.5v^2 \rightarrow h=(0.5v^2)/g \rightarrow h=(0.5 \times 1^2)/9.81 \rightarrow h = 0.5/9.81 = 0.0509683 \Rightarrow 0.051 metres$

-final mass is ten times, so final velocity will be one tenth (by conservation of momentum) vf = 1m/s

-1/2 m v^2 = mgh => gh=0.5 =>h= 0.051

Here, since it is said to be an inelastic collision, we initially cannot conserve energy while calculating the speed of the pendulum just after the hit because some amount of energy will be required in holding the 2 bodies together. Thus we will need to conserve momentum. Let the final speed of both the pendulum and the gum be ' v m/s ' (inelastic_therefore both the bodies will have the same speeds after the hit).

Thus we get:
Initial momentum of the gum=final momentum of the gum+final momentum of the pendulum

=> (10m/s)(10g)=(10g)v+(90g)v

We get v=1m/s. Now we will conserve energy between this point and the point where the pendulum stops momentarily after attaining a height of 'h meters'.

We get:
Initial Kinetic Energy=Final Potential Energy

(m)(v)(v)/2=m g h....................................(mass m= 90+10)(v=1)

=> (100)/2=100(9.8)h

We get h=0.051020408163 meters after solving the above equation

During inelastic collision, energy is not conserved. But momentum is. Let $m_1$ be mass of gum, $m_2$ , mass of pendulum, $v_i$ initial speed of gum, $v_f$ final speed of new pendulum.

$m_1v_1= (m_1 + m_2)v_f \Rightarrow v_f=1m/s$

Now we apply conservation of energy to the pendulum and it reaches maximum height when it exhausts all its kinetic energy.

$\frac{(m_1 + m_2) v_f^2}{2}=(m_1 + m_2)gh_{max} \Rightarrow h_{max}=0.051m/s$

Because momentum is conserved:

$p=mv$

$p_{initial}=0.01 \times 10 = .1$

$m_{final}=.09+.01=.1$ kg

$.1=.1v$

$v_{final}=1$ m/s

The kinetic energy will be equal to the gravitational potential energy:

$\frac{1}{2}mv^2=mgh$

$\frac{1}{2}v^2=gh$

$h=\frac{v^2}{2g}=\frac{1}{19.6}=0.0510$

This is a simple problem on ballistic pendulum. Here, an inelastic collision takes place between the gum and the pendulum. Since, deriving the general equation helps a lot in understanding relation between different variables I will take the same path. Let the mass of the pendulum be m* and that of gum be m. Let the velocity of the gum be v. By the law of conservation of momentum,

                initial momentum=final momentum

         mv + m*(0)=(m + m*)v*         where v* is the velocity gained by the pendulum after the collision.
                   We get,  

                            v*=mv/(m + m*)

Thus the pendulum gains kinetic energy and this energy gets converted to its potential energy as it rises up. By the law of conservation of energy, kinetic energy=potential energy at max. height

We get,

                              h= (v*)^2/(2g)

                               =   (mv)^2/(m+m*)^2*(2g)

Plug in the values and get the answer.

From the Principle of Conservation of Linear Momentum,

$m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$

where $m_1 = 10g, m_2 = 90g, u_1 = 10m/s, u_2 = 0, v=?$

Solve for v and we get $v = 1 m/s$

Then, by Principle of Conservation of Energy,

$\frac{1}{2} mv^2 = mgh$

$h = \frac{ v^2 }{2g}$

$h= 0.051 m$

We consider the different components of the problem.

Part 1 : The collision. If we work with mass in grams and velocities in m/s, and if the mass of the gum and the pendulum bob are $m$ and $M$ , respectively, and $h$ is the maximum height, and $v_0 = 10 \text { m/s}$ , we have $mv_0 = (m + M)v,$ where $v_0$ is the speed of the sticky pendulum bob after the collision (with the gum sticking to it). Since $m + M = 10m$ , we have $v = \dfrac {v_0}{10}$ .

Part 2 : Now that we have the take-off speed of the sticky system, we can use Conservation of Energy to find the maximum height of the bob. We have that the energy at the bottom is all kinetic: $\dfrac {1}{2} (m + M)v^2,$ and the energy at the tom is all potential: $(m+M) gh.$ Thus, setting these equal, we get $v^2 = 2gh$ , so $h = \dfrac {v^2}{2g}$ .

Since $v = \dfrac {v_0}{10} = \dfrac {10 \ \text{m/s}}{10} = 1 \ \text{m/s}$ , we have $h = \left( \dfrac {1}{19.6} \right) \ \text{m} = \boxed {5.1 \times 10^{-2} \ \text{m}}.$

This is a conservation of momentum and energy problem.

For the conservation of momentum, we have:

$m_{gum})(v_{gum}) + (m_{pendulum})(v_{pendulum}) = (m_{gum}+m_{pendulum})(v_{system})$

Plugging all the values we have:

$10g)(10\frac{m}{s}) +(90g)(0\frac{m}{s}) = (100g)(v_{system})$ $v_{system} = 1\frac{m}{s}$

For the conservation of energy we have:

$mgh_{i} + \frac{1}{2}(mv_{i}^{2}) = mgh_{f} + \frac{1}{2}(mv_{f}^{2})$ .

Note that we are looking for $h_{f}$ , so $h_{i} =0$ and $v_{f} =0$ .

Therefore $\frac{1}{2}(mv_{i}^{2}) = mgh_{f}$ . Note that $m$ will cancel. Therefore:

$\frac{1}{2}(v_{i}^{2}) = gh_{f} \Rightarrow \frac{\frac{1}{2}(1\frac{m}{s})^{2}}{g} = h_{f}$

Then $h_{f} = \frac{1}{2 * 9.8} = \frac{1}{19.6} = 0.051$

By principle of conservation of momentum, $(10)(10)+(90)(0)=(100)(v)$ , $v=1m/s$ . By principle of conservation of energy, $mgh=\frac{1}{2}mv^2$ , $(m)(9.8)(h)=\frac{1}{2}(m)(1)^2$ , $h=0.051m$ .

We use the Law of Conservation of Linear Momentum to determine the final velocity of the system after collision.

We know that:

velocity of gum = v_g = 10 m/s

velocity of the pendulum bob = v_p = 0 m/s

mass of gum = m_g =10 g

mass of pendulum = m_p = 90 g

We will look for:

final velocity = v_f

By the Law of Conservation of Linear Momentum, we get:

(10 g)(10 m/s) + (90 g)(0 m/s) = (100 g)(v_f)

100 = 100(v_f)

v_f = 1 m/s

Now we know from the Law of Conservation of Energy, that with respect to an arbitrary point:

v_f = \sqrt{2gh}

Substituting v_f, we get:

1 m/s = \sqrt{2(9.8 m/s)(h)}

Squaring both sides:

1 = 2(9.8)h

1 = 19.6 h

h = 0.051 m