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3 years ago

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Two ice skaters stand facing each other at rest on a frozen pond. They push off against one another and the 48 kg skater acquire

s a speed of 0.67 m/s. If the other skater acquires a speed of 0.83 m/s, what is her mass in kilograms?

2 answers:

BaLLatris [955]3 years ago

5 0

Answer:

The mass from the second skater is 40.2Kg

Explanation:

To solve this problem it is necessary to apply the concepts related to conservation of the moment.

By definition for the particular case we have to,

$m_1u_1 = m_2v_2$

Where,

$m_1$ = Mass first skater

$u_1$ = Speed of first skater

$v_2$ = Velocity of second skater

Our values are given as,

$m_1= 48Kgu_1 = 0.67m/sv_2 = 0.8m/s$

Replacing in our equation and re-arrange to find $m_2,$

$m_2 = \frac{m_1u_1}{v_2}m_2 = \frac{(48)(0.67)}{0.8}m_2 = 40.2Kg$

The mass from the second skater is 40.2Kg

Papessa [141]3 years ago

5 0

Answer:

38.75 kg

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = mv+m'v'................... Equation 1

Note: Since both skater were initially at rest, u = u' = 0 m/s.

Also Note that both skater moves in opposite direction after collision, and let the direction of the first skater be positive and that of the second skater be negative.

mv = -m'v'.......................... Equation 2

make m' the subject of the equation above,

m' = -mv/v'..................... Equation 3

Given: m =48 kg, v = 0.67 m/s, v = -0.83 m/s (opposite direction to the first skater)

substitute into equation 3

m' = -48(0.67)/-0.83

m' = -32.16/-0.83

m' = 38.75 kg

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3 years ago