# A 200 g puck sliding on ice strikes a barrier at an angle of 53 degrees and it bounces off at an...

## Question:

A 200 g puck sliding on ice strikes a barrier at an angle of 53 degrees and it bounces off at an angle of 45 degrees. The speed of the puck before the bounce was 15 m/s and after is 12 m/s. Changes in direction require force.

a. Was energy lost in the bounce? How do you know?

b. If the impulsive force only had the normal component and the friction component are zero. What should be the final velocity of the puck?

## Conservation of energy:

The energy of the system remains conserved when there is a change or disturbance in the system. One form of energy converts to another in the form of heat. The transformation of the energy to heat is also referred conversion to heat energy.

Given data

• The value of the mass of the puck is {eq}{m_p} = 200\;{\rm{gm}} = 0.2\;{\rm{kg}} {/eq}
• The value of the angle {eq}\phi = 53^\circ {/eq}
• The value of the angle {eq}\theta = 45^\circ {/eq}
• The value of the initial speed of the puck is {eq}{v_i} = 15\;{\rm{m/s}} {/eq}
• The value of the final speed of the puck is {eq}{v_f} = 12\;{\rm{m/s}} {/eq}

(a)

The expression for the loss of energy is,

{eq}\begin{align*} {E_{loss}} &= K.{E_{initial}} - K.{E_{final}}\\ &= \dfrac{1}{2}{m_p}{v_i}^2 - \dfrac{1}{2}{m_p}{v_f}^2 \end{align*} {/eq}

Substitute the value in the above equation.

{eq}\begin{align*} {E_{loss}} &= \dfrac{1}{2}\left( {0.2} \right){15^2} - \dfrac{1}{2}\left( {0.2} \right){12^2}\\ &= 8.1\,{\rm{J}} \end{align*} {/eq}

Thus, the value in the loss of energy in puck is {eq}8.1\;{\rm{J}} {/eq}

(b)

The expression for the final velocity of the puck in x direction,

{eq}{v_{{f_{x\;axis}}}} = {v_f}\cos \theta {/eq}

Substitute the value in the above equation.

{eq}\begin{align*} {v_{{f_{x\;axis}}}} &= 12\cos 45^\circ \\ &= 8.4852\;{\rm{m/s}} \end{align*} {/eq}

The expression for the final velocity of the puck in y direction,

{eq}{v_{{f_{y\;axis}}}} = {v_f}\sin \theta {/eq}

Substitute the value in the above equation.

{eq}\begin{align*} {v_{{f_{x\;axis}}}} &= 12\sin 45^\circ \\ &= 8.4852\;{\rm{m/s}} \end{align*} {/eq}

The expression for the resultant final velocity of the puck is,

{eq}{v_{ne{t_{final}}}} = \sqrt {{v_{{f_{x\;axis}}}}^2 + {v_{{f_{y\;axis}}}}^2} {/eq}

Substitute the value in the above equation.

{eq}\begin{align*} {v_{ne{t_{final}}}} &= \sqrt {{{\left( {8.48} \right)}^2} + {{\left( {8.48} \right)}^2}} \\ &= 8.48\;{\rm{m/s}} \end{align*} {/eq}

Thus, the value of the final velocity of the puck is {eq}8.48\;{\rm{m/s}} {/eq} 