Complete the following problems, showing all calculations on which you base your

Question

Complete the following problems, showing all calculations on which you base your answers. Put a circle (box) around your answers. Students are expected to work independently on the problems. Use -9.8 m/s^2 as the value for the acceleration due to gravity. Three very young ice skaters collide, as shown below, and stick together. What is the velocity of combined mass at the instant after impact? How far will their combined masses travel across the ice (with coefficient of friction = 0.07), before they come to a stop?

Explanation / Answer

According to the law of conservation of linear momentum, the net linear momentum should be conserved before and after the collision. Moreover it should be conserved in all the directions.

Here let us work in the x and y directions. For thast we shall the velocities Va, Vb and Vc into x and y components.

In the X-direction:

Net momentum before collision = Net momentum after collision

ma x (-VaCos20) + mb x (VbCos48) + mc x (VcSin 31) = M Vx

18x(- 5.2 Cos 20) +20 x (9.7Cos48) + 21 x (7.5 Sin 31) = (18 +20 + 21) Vx

Vx = 2.084315341 m/s

In the Y-direction:

Net momentum before collision = Net momentum after collision

ma x (VaSin20) + mb x (-VbSin48) + mc x (VcCos 31) = M Vy

18x(5.2 Sin 20) +20 x (- 9.7Sin48) + 21 x (7.5 Cos 31) = (18 +20 + 21) Vy

Vy = 0.387234561 m/s

Vx = 2.084315 m/s

So the net velocity of the combined mass after the collision is V = ( Vx2 + Vy2 )1/2 = 2.119981379 m/s

Frictional force on combined mass M is Mgk

retardation due to this frictional force is = Mgk / M = gk = -9.8 x 0.07 = - 0.686 m/s2 the negative sign just symbolizes the retardation.

using the formula v2 - u2 = 2as

02 - 2.1199813792 = 2 (-0.686) s

s = 3.275744204 m

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