Question

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

Answer 1

PART a)

Momentum is defined as product of mass and velocity

so here initial momentum is

$P = mv$

given that

m = 0.3 kg

v = 18 m/s

$P = 0.3(18) = 5.4 kg m/s$

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

$F = \frac{\Delta P}{\Delta t}$

so here we have

time interval = 0.25 s

now from above equation

$F = \frac{5.4 - 0}{0.25}$

$F = 21.6 N$

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

$Impulse = P_f - P_i$

$impulse = 3.9 - 5.4 = -1.5 kg m/s$

Part d)

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

Initial momentum of the puck is given as

$P = mv = (0.3)(13) = 3.9 kg m/s$

so total momentum before collision is given as

$P = 3.9 + 0 = 3.9 kg m/s$

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

$P_i = P_f$

$3.9 = (m_1 + m_2) v$

$3.9 = (0.3 + 1.2)v$

$v = 2.6 m/s$

so final combined speed will be 2.6 m/s

Answer 2

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