robert7248
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Alan hits a hockey pick of mass 0.3 kg across a hockey rink with a velocity of 18 m/s to the east.

a. What is the momentum of the hockey puck?

b. If the puck was initially at rest and the time of impact was 0.25 seconds, what force did Alan apply to the puck when he hit it?

c. The puck slides across the rink, losing 5 m/s of speed due to friction. What impulse is applied by friction?

d. When Alan's puck is moving at 13 m/s east, it collides with a sticky foam block of mass 1.2 kg sitting on the rink. The puck sticks in the foam block and they move together. What was the momentum of the pick-block system before the collision?

e. What is the speed of the puck-block system after the collision?

Respuesta :

PART a)

Momentum is defined as product of mass and velocity

so here initial momentum is

[tex]P = mv[/tex]

given that

m = 0.3 kg

v = 18 m/s

[tex]P = 0.3(18) = 5.4 kg m/s[/tex]

towards East

Part b)

Since puck was initially at rest so here initial momentum must be ZERO

so here to find the force we can use

[tex]F = \frac{\Delta P}{\Delta t}[/tex]

so here we have

time interval = 0.25 s

now from above equation

[tex]F = \frac{5.4 - 0}{0.25}[/tex]

[tex]F = 21.6 N[/tex]

Part c)

Due to friction the puck lose its speed by 5 m/s

final speed = 18 - 5 = 13 m/s

mass = 0.3 kg

final momentum = (0.3)(13) = 3.9 kg m/s

now the impulse due to friction force is given as

[tex]Impulse = P_f - P_i[/tex]

[tex]impulse = 3.9 - 5.4 = -1.5 kg m/s[/tex]

Part d)

initial momentum of block will be ZERO as it is placed at rest

Initial momentum of the puck is given as

[tex]P = mv = (0.3)(13) = 3.9 kg m/s[/tex]

so total momentum before collision is given as

[tex]P = 3.9 + 0 = 3.9 kg m/s[/tex]

Part e)

since the system is isolated and there is no external force on it

So here momentum will remain conserved

so here we have

[tex]P_i = P_f[/tex]

[tex]3.9 = (m_1 + m_2) v[/tex]

[tex]3.9 = (0.3 + 1.2)v[/tex]

[tex]v = 2.6 m/s[/tex]

so final combined speed will be 2.6 m/s

a) p=mv=5.4 kg*m/s

b) F=p/t=21.6 N

c) p=0.3*13-p=-1.5 kg*m/s

d) According to law of conservation of momentum, p.b=0.3*13=3.9 kg*m/s

e) Apply the same law: v=p.b/(0.3+1.2)=3.9/(0.3+1.2)=2.6 kg*m/s

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