Question

Problem 2.14 Your answer is partially correct. Try again A 0.52 kg block of ice is sliding by you on a very slippery floor at 2.4 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 0.0022 seconds. The bick eventually slides at an angleof 29 degrees from its original direction labeled θ in the d agram). The overhead view shown in the diagram is a d ately to scale. The a w re esent the average force your toe applies briefly to the block of ice (a) Which of the possible paths shown in the diagram corresponds to the correct overhead view of the bleck's path? b) 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 1Z plane.) z component x component y component e What 1s the unit vector in the direction of the block's momentum after the kick? 庐= 087 ia (d) What is the x component of the block's momentum after the kick?

Answer 1

If we draw a right angle triangle with velocity in x direction ( $v_x$ ) as the base then velocity in z direction ( $v_z$ ) is the perpendicular side of $v_x$ . The angle between  and the hypotenuse is $\theta=29^o$

(f) Now, $tan\,\, \theta=\frac{v_z}{v_x}$

$\\\\v_z=v_x\,\,tan\,\, \theta\\\\=2.4\times tan\,\,29\\\\=1.33\,\,m/s$

Then, z component of the block's momentum

  $\\\\P_f_z= m\times v_z=0.52\times 1.33= 0.692\,\,kgm/s$

(e) Magnitude of the momentum after kick

  $\\\\P=\sqrt(P_f_x+P_f_z)\\\\=\sqrt(1.25^2+0.692^2)\\\\=1.429\,\,kgm/s$

​​​​​(g) Magnitude of average force applied to the block,

$F_a_v=\frac{P_f_z}{t}$

$=\frac{0.692}{0.0022}$

$=314.5=315 \,\,N$