Question

A 0.17 kg hockey puck slows down trom 54 m/s to 35 m/s when it slides on horizontal ice surface. Find the change in momentum of the puck?

Answer 1

Answer:

D_momentum = 3.23[kg*m/s]

Explanation:

The change of moment can be easily determined by applying the principle of momentum and amount of movement

initial data are:

$m_{1}=0.17[kg]\\ m_{2}=0.17[kg]\\v_{1}=54[m/s]\\v_{2}=35[m/s]\\D_{momentum}= m_{1}*v_{1}-m_{2}*v_{2}\\D_{momentum}=0.17(54-35)\\D_{momentum}= 3.23[kg*m/s]$

Answer 2

The change in momentum will be "3.23 kg.m/s".

Given values,

Mass,

• $m_1 = 0.17 \ kg$
• $m_2 = 0.17 \ kg$

Velocity,

• $v_1 = 54 \ m/s$
• $v_2 = 35 \ m/s$

We know,

→ $D_{momentum } = m_1\times v_1-m_2\times v_2$

By putting the values,

                     $= 0.17\times 54-0.17\times 35$

                     $= 0.17(54-35)$

                     $= 0.17\times 19$

                     $= 3.23 \ kg.m/s$

Thus the above approach is right.                

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