Question

A meter stick is at rest on frictionless surface. A hockey puck is going towards the 30cm mark on the stick and is traveling perpendicular to the stick. After the collision the puck is deflected 30 degrees from original path and is traveling half its original speed.
~Mass of a meter stick = 0.05 kg~Mass of hockey puck = 0.17 kg~Initial speed of the hockey puck = 9 m/sA) Choosing an origin at the starting position of the meter stick's center of mass, what is the angular momentum of the hockey puck before the collision.B) What is the angular momentum of the hockey puck after the collision? (Use same origin.)C) What is the velocity (direction and speed) of the stick's center of mass after the collision?D) What is the angular velocity of the stick (assume it will rotate about its center of mass)?

Answer 1

Answer:

A)$0.306k$

B)$0.1325k$

C)$v_{s_{f}} =7.65(-i)+17.35(-j)$

D)$w=41.64 rads^{-1}$

Explanation:

Given:

• hockey puck is moving towards 30cm mark perpendicular to the stick
• $m_{s} = 0.05kg$
• $m_{h} =0.17kg$
• $v_{h_{i} } = 9 ms^{-1}$
• after collision the puck is deflected 30°
• $v_{h_{f} } =4.5 ms^{-1}$

To find the initial angular momentum about origin which is the 50th mark of the metre scale (It's COM) :                                                                                                                          angular momentum $L$=$mv$x$r$                                      where,                                                                                                                       v  - is the velocity of the puck perpendicular to the radial vector

r  -  is the radius vector

∴

A) $L_{i} = m_{h}r*v_{h_{i} }\\L_{i}= 0.17*\frac{50-30}{100} (-i)*9(-j)\\L_{i} = 0.306 k$

B) after collision , it moves 30° from original path;

   and it's speed = $\frac{9}{2} =4.5ms^{-1}$;

   ∴the perpendicular velocity $v_{per}$ = 4.5$cos30$ = $2.25\sqrt{3}$$ms^{-1}$

⇒$L=m_{h}r*v=0.17*0.2(-i)*2.25\sqrt{3}(-j) \\L= 0.1325 k$

C) since the net external force on the system is zero , the total momentum of the system can be conserved .

thus ,

$m_{s}v_{s_{i}}+m_{h}v_{h_{i}}=m_{s}v_{s_{f}}+m_hv_{h_{f}}\\0+0.17*9(-j)=0.05*v_{s_{f}}+0.17*(2.25(i)+2.25\sqrt{3}(-j))$

solving this we get,

⇒$v_{s_{f}} =7.65(-i)+17.35(-j)$

D) since there is no external torque about the system ,the angular momentum can be conserved.

$L_{h_{i}}= L_{h_{f}} + Iw$

where ,

$w$ is the angular velocity of the stick.

$I$ is the moment of inertia of the stick about COM :

$I =\frac{m_{s}l^{2}}{12} \\m_{s}=0.05kg\\l=1m\\I = 0.004167 kgm^{2}$

∴

⇒$0.306k=0.1325k+0.004167w\\w=41.64 rads^{-1}$