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a) Julie is running to the right at 5 m/s, as shown in the 1st figure. Balls 1 and 2 are thrown toward her at 10 m/s by friends standing on the ground. According to Anita, what's the speed of the two balls?

b) Julie is running to the right at 5 m/s, as shown in the 2nd figure. Balls 1 and 2 are thrown toward her by friends standing on the ground. According to Julie, both balls are approaching her at 10 m/s. According to Julie's friends, what's the speed of the two balls?

Please give explanations and step-to-step solution and calculus. Thank you!

a Julie is running to the right at 5 ms as shown in the 1st figure Balls 1 and 2 are thrown toward her at 10 ms by friends standing on the ground According to A class=
a Julie is running to the right at 5 ms as shown in the 1st figure Balls 1 and 2 are thrown toward her at 10 ms by friends standing on the ground According to A class=

Respuesta :

Part 1)

here Julie is running at speed 5 m/s

So here two balls are thrown at speed 10 m/s towards Julie with respect to her friend standing on the ground.

So here this all speed is real speed of all.

Now as per Anita (let say she is one of her friend standing on ground) the speed of two balls will be same as the given speed as she is observing from ground or stationary frame

As per the frame of Julie

speed of ball 1

[tex]v_{1j} = v_1 - v_j[/tex]

[tex]v_{1j} = 10 - 5 = 5 m/s[/tex] towards her in same side

speed of ball 2

[tex]v_{2j} = v_2 - v_j[/tex]

[tex]v_{2j} = -10 - 5 = -15 m/s[/tex] towards her from opposite side

Part b)

Now in this case the speed of two balls is given with respect to Julie

so we can say

for ball 1

[tex]v_{1j} = v_1 - v_j[/tex]

[tex]10 = v_1 - 5 [/tex]

so in ground frame speed of ball 1 is

[tex]v_1 = 15 m/s[/tex]

Similarly for ball 2

[tex]v_{2j} = v_2 - v_j[/tex]

[tex]-10 = v_2 - 5[/tex]

So speed in ground frame of ball 2 is

[tex]v_2 = - 5 m/s[/tex]

so its 5 m/s from opposite side

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