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

A) The angular momentum of the hockey puck before the collision is 0.459 kg.m^2/s. B) The angular velocity of the puck after the collision is 15 rad/s. C) The velocity of the stick's center of mass after the collision is 3.475 m/s.

Step-by-step explanation:

A) The angular momentum of an object is given by the equation:

L = Iω

where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

Since the meter stick is at rest, its angular velocity is zero, and therefore its angular momentum is also zero.

On the other hand, the angular momentum of the hockey puck before the collision can be calculated using the equation:

L = mvr

where m is the mass of the puck, v is its linear velocity, and r is the distance from its axis of rotation.

Using the given values, the angular momentum of the hockey puck before the collision is 0.17 kg * 9 m/s * 0.3m = 0.459 kg.m^2/s.

B) After the collision, the puck is deflected at an angle of 30 degrees and its linear velocity is halved.

The angular velocity of the puck after the collision can be calculated using the equation:

ω = v'/r'

where v' is the new linear velocity of the puck and r' is the new distance from its axis of rotation.

Since the linear velocity of the puck is halved, its new linear velocity is 4.5 m/s.

Assuming the radius of rotation remains the same, the angular velocity of the puck after the collision is 4.5 m/s / 0.3m = 15 rad/s.

C) The velocity of the stick's center of mass after the collision can be calculated using the conservation of linear momentum.

Before the collision, the momentum of the system is zero, since the meter stick is at rest.

After the collision, the momentum of the system is the linear momentum of the puck, which can be calculated using the equation:

p = mv

where m is the mass of the puck and v is its linear velocity.

Using the given values, the linear momentum of the puck after the collision is 0.17 kg * 4.5 m/s = 0.765 kg.m/s.

Since the meter stick and the puck move together after the collision, their velocities are the same and equal to 0.765 kg.m/s divided by the total mass of the system, which is 0.05 kg + 0.17 kg = 0.22 kg.

Therefore, the velocity of the stick's center of mass after the collision is 0.765 kg.m/s / 0.22 kg = 3.475 m/s.

D) The angular velocity of the stick can be calculated using the equation:

ω = v/r

where v is the linear velocity of the stick's center of mass and r is the radius of the stick.

Using the given values, the linear velocity of the stick's center of mass after the collision is 3.475 m/s.

Assuming the radius remains the same, the angular velocity of the stick is 3.475 m/s / 0.3m = 11.583 rad/s.

Answer 2

A)
`0.306k`

B)
`0.1325k`

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

D)
`w=41.64 rads^(-1)`

Step-by-step 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*(50-30)/(100) (-i)*9(-j)\\L_(i) = 0.306 k`

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

and it's speed =
`(9)/(2) =4.5ms^(-1)`;

∴the perpendicular velocity
`v_(per)` = 4.5
`cos30` =
`2.25√(3)`
`ms^(-1)`

⇒
`L=m_(h)r*v=0.17*0.2(-i)*2.25√(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√(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 =(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)`