A 06 kg block of Ice is sliding by you on a very slippery fl

A 0.6 kg block of Ice is sliding by you on a very slippery floor at 3 m/s. As it goes by, you give it a kick perpendicular to its path. Your foot is in contact with the ice Nock for 0.002S seconds. The block eventually sides at an angle of 24 degrees from its original direction (labeled theta in the diagram). The overhead view shown in the diagram is approximately to scale. The arrow represents the average force your ice applies briefly to the bock of ice. Which of the possible paths shown in the diagram corresponds to the correct overhead view of the block\'s path? Which components of the block\'s momentum are changed by the impose applied by your foot? What a the unit vector in the direction of the block\'s momentum after the kick? What n the x-component of the block\'s momentum after the kick? What is the magnitude of the block\'s momentum after the kick? Use your answers to the preceding questions to find the z-component of the block\'s momentum after the kick (drawing a diagram is helpful): What was the magnitude of the average force you applied to the block?

Solution

path r is related as with force.

d^2 z / dt^2 = F/m

so it will be figure C.

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force is along z axis, so only z component will change.

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unit vector in the drection of momentum after kick

= 1 ( cos24i - sin24j) = 0.913i - 0.407j

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x -compoennt will be unchanges

final Px = initial px = mv = 0.6 x 3 = 1.8 kg m/s

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x component is px = pcos24

so p = 1.8 / cos24 = 1.97 kg m/s

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pz = psin24 = 1.97 sin24 = 0.801 kg m/s

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pz = F * deltat

0.801 = F * 0.0025

F = 320.56 N