can someone please do question no.1 both parts on paper and can you please show

Question

can someone please do question no.1 both parts on paper and can you please show work (explain).Thank you

Date: PRoeLEM Two ice skaters, Daniel (mass 6g.2 kg) and Rebecca 50.8 kg) are practicing. Daniel stops to tie his After and, while at rest, is struck by Rebecca, who is moving at 13.2 s before she collides with him. the collision, Rebecca has a direction. Both skaters move on the frictionless, horizontal surface of the rink. 53.75 from her initial magnitude 6.3 angle of (a) What are the magnitude and direction of Daniers velodty arter the collision? (b) What is the change in total kinetic energy of the two skaters as a result of the collision?

Explanation / Answer

Let v1i and v1f be Daniel's velocity before and after the collision respectively who’s mass is m1.

Let v2i and v2f be Rebecca's velocity before and after the collision respectively who’s mass is m2.

Let Rebecca is initially travelling in positive x-direction.

Then by conservation of linear momentum

Along the horizontal component, (let Rebecca is initially moving along positive x direction)

m1 * v1i + m2 * v2i = m1 * v1f cos 1+ m2 * v2f cos 2     ………………………. (1)

We know that, before collision Daniel’s velocity is zero.

Hence from (1)

0 + (50.8 * 13.2) = (69.2 * v1f cos 1) + (50.8 * 6.3 * cos 53.7º)

v1f cos 1                   = 6.95   …………… (2)

Along the vertical component, (since Rebecca is initially moving along x –direction, y component or vertical component of Rebecca is zero initially and also Daniel initial velocity is zero. )

Hence

m1 * v1i + m2 * v2i = m1 * v1f sin 1+ m2 * v2f sin 2     

0 = - (69.2 * v1f sin 1) + (50.8 * 6.3 * sin 53.7º)

(69.2 * v1f sin 1) = (50.8 * 6.3 * sin 53.7º)  
                                         [ Both the angles are in opposite direction, and 1 will have a   

                                            negative sign with respect to rebeca’s initial direction ]

v1f sin 1 = 3.73      ……………………..(3)

Squaring and adding (2) and (3)

v1f = 7.88 m/s

Hence solving (2) or (3) we have

1 = 28 º

Change in kinetic energy is given by

= ½ m2 (v2f2 –v2i2) + ½ m1 (v1f2 –v1i2)

Substituting all the values we have

= - 1269.11 J ( the negative sign indicates that total kinetic energy has decreased during the process)

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