A 0.6 kg block of ice is sliding by you on a very slippery floor at 4 m/s. As it

Question

A 0.6 kg block of ice is sliding by you on a very slippery floor at 4 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 0.003 seconds. The block eventually slides at an angle of 26 degreerees 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 toe applies briefly to the block 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 impulse applied by your foot? (Check all that apply. The diagram shows a top view, looking down on the xz plane.) z component y component x component What is the unit vector in the direction of the block's momentum after the kick? P = What is the x-component of the block's momentum after the kick? Pfx = kg . m/s Remember that p rightarrow= |p rightarrow |p. What is the magnitude of the block's momentum after the kick? |p rightarrow | = kg . m/s 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): Pfz = kg . m/s What was the magnitude of the average force you applied to the block? |F rightarrow avgl = N

Explanation / Answer

As there is no momentum change in x direction and in z direction there is a velocity due to the foot by after it leaves the contact of foot no acc in z direction. So no acceleration in x and no acceleration z therefore path is B Z component of momentum is changed as force applied is also in Z direction and as according to newtons 2nd law change in momentum in that direction = net impulse in same direction. As angle is 26 therefore vertical velocity vy = 4 * tan(26) therefore unit vector of direction = [ i+tan(26)j ]/[squareroot(1+(tan(26))^2] X component of momentum is = same as initial therefore it is 4*.6 = 2.4 kg m/s (no external force acting in x) momentum after kick is = .6*(4i + 4*tan(26)j) magn = 2.67 Z component of momentum = 2.4*tan(26) avg force = 2.4*tan(26) / .003 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.