A 0.55 kg block of ice is sliding by you on a very slippery floor at 3.5 m/s. As
ID: 3307581 • Letter: A
Question
A 0.55 kg block of ice is sliding by you on a very slippery floor at 3.5 m/s. As it goes by, you give it a kick perpendicular to its path. Your foot is in contact with the ice block for 0.002 seconds. The block eventually slides at an angle of 26 degrees from its original direction labeled in the diagram The overhead view shown in the diagram is appro inte r to sal The arrow represents the average force your toe applies briefly to the block of ice Which Selec- Which components of the block's momentum are changed by the impulse applied by your foot? (Check all that apply. The diagram shows a top view, looking down on the xz plane.) z component
Explanation / Answer
Figure C is the correct overhead view of the block's path.
Figure A and B are incorrect because a force only can apply acceleration which leads to change its velocity gradually in the direction of the applied force and hence the displacement in that direction during the application of force should increase gradually which is correctly depicted in figure C.
Z-component of the momentum is changed because the force is applied in negative Z-direction only.
cos26oi+sin26ok is the unit vector in the direction of the block's momentum after the kick where i and k are unit vectors in the direction of x and z respectively.
X-component remains same which is equal to mv = .55*3.5 =1.925 kg-m/s
We have Pcos26o =1.925 which gives
P=1.925/cos26o
P=1.925/.899 = 2.14 kg-m/s
P=2.14 kg-m/s
Z-component of the momentum is Psin26o =2.14*.438 = .938 kg-m/s
Force applied is equal to change in momentum per unit time in the direction of applied force.
So change in the Z-component of the momentum is .938 kg-m/s (As it was zero initially).
The time during the force is applied = .002 s.
F=.938/.002 = 469 kg-m/s2 =469 N
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