You (m = 50 kg) decide to go skating with your friend (m = 65 kg). At some point

Question

You (m = 50 kg) decide to go skating with your friend (m = 65 kg). At some point, you are skating at 4.0 m/s (in -y-direction) when suddenly your friend is coming from the side(-x direction) and collides with you. During the collision you hold on to each other, but you both fall and slide together over the ice for 3.5 m. If you slide at an angle alpha = 55 degrees with respect to your initial direction, what was your friend's speed? What's the coefficient of kinetic friction in this case? An object at rest explodes into three fragments. What is the speed of the third fragment that has a mass of 125 g?

Explanation / Answer

1)
let

m1 = 50 kg, u1 = 4 m/s

m2 = 65 kg, u2 = ?

Let V is the spped of the two persons after the collision.

Apply conservation of momentum in y-direction

m1*u1 = (m1+m2)*v*cos(55)

50*4 = (50+65)*V*cos(55)

==> V*cos(55) = 200/115

V*cos(55) = 200/115 ---(1)

Apply conservation of momentum in x-direction

m2*u2 = (m1+m2)*v*sin(55)

65*u2 = (50 + 65)*v*sin(55)

v*sin(55) = 65*u2/115   ---(2)

take equation(2)/equation(1)

tan(55) = 65*u2/200

==> u2 = 200*tan(55)/65

= 4.4 m/s <<<<<<<<<<---------------------Answer

2)

after the collsion, the speed of two persons,
v = 200/(115*cos(55))

= 3.03 m/s

acceleration, a = (vf^2 - vi^2)/(2*d)

= (0^2 - 3.03^2)/(2*3.5)

= -1.3

we know, a = -g*mue_k

==
> mue_k = -a/g

= 1.3/9.8

= 0.1326 <<<<<<<<---------Answer

2) In any explosion Net momentum must be equal to zero,Pnet = 0

P1 + P2 + P3 = 0

P3 = -(P1 + P2)

= -( (-2i + 2j) + (2i) )

= -2 j kg.m/s

P3 = -2 kg m/s

m3*v3 = 2

v3 = 2/m3

= 2/0.125

= 16 m/s <<<<<<---------Answer

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