Question

A 60-kg skater pushes a 50 kg skater, who moves away at
2.0 m/s. As a result, the first skater moves backward at
(a) 0.6 m/s (b) 1.7 m/s (c) 2.0 m/s (d) 2.4 m/s​

Answer 1

Answer:

(b) 1.7 m/s is the correct answer

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Answer 2

Answer:

(b)The first skater moves backward with velocity of 1.7m/s

Explanation:

Given that,

The mass of First skater is

$m_{1} = 60kg$

The mass of second skater is

$m_{2} = 50kg$

The initial velocities if both the skaters is zero.Therefore,

$u_{1} = u_{2} = 0$

Now the final velocity of second skater is given in the question which is 2m/s.Hence,

$v_{2} = 2ms {}^{ - 1}$

According to law of conservation of momentum,

The initial momentum of both the skater is equal to their final momentum.Therefore,

$m_{1}u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$

Substituting given values,we get

$0 = 60v_{1} + 50(2) \\ = > v_{1} = \frac{ - 100}{60} = - 1.67ms {}^{ - 1}$

Negative sign indicates that the skater moves in opposite direction.

Therefore,The first skater moves backward with velocity of 1.7m/s(near to given options)

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