A 3.40 kg steel ball strikes a wall with a speed of 10.0 m/s at an angle of 60.0
ID: 2142435 • Letter: A
Question
A 3.40 kg steel ball strikes a wall with a speed of 10.0 m/s at an angle of 60.0
A 3.40 kg steel ball strikes a wall with a speed of 10.0 m/s at an angle of 60.0 degree with the surface. It bounces off with the same speed and angle (Fig. P6.14). If the ball is in contact with the wall for 0.225 s, what is the average force exerted on the ball by the wall? A 755 N man stands in the middle of a frozen pond of radius 6.0 m. He is unable to get to the other side because of a lack of friction between his shoes and the ice. To overcome this difficulty, he throws his 1.2 kg physics textbook horizontally toward the north shore at a speed of 6.0 m/s. How long does it take him to reach the south shore?Explanation / Answer
Fy = dpy/dt
Fx = dpx/dt
dp =change in momentunm...
Px =change in momemtum along x direction
Py=along y direction
Fx =force along x direction
Fy = 0.
Fx = dpx/dt = px / t
px,i = m vx,i = (3.4 kg) (10 m/s) sin 60
sin 60 = 0.866
px,i = (3.4 kg) (10 m/s) (0.866) = 29.444 kg m/s
px,f = m vx,f = - (3.4 kg) (10 m/s) sin 60
px,f = - (3.30 kg) (10 m/s) (0.866) = - 29.444 kg m/s
px = px,f - px,i
px = - 29.444 kg m/s - 24.444 kg m/s
px = - 58.888 kg m/s
Fx = px / t
Fx = -58.888/0.225
Fx = - 261.7244
since there is no change ion velocity along x
Fy = 0
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Apply conservation of momentum. When he throws the textbook, he will get a push from the reaction force of the textbook on him.
mbvb = -mmvm
vm = -mbvb/mm = -(1.2kg)(6.0m/s)/((755N)/(9.8m/s2)) = -0.0934 m/s. (minus sign means the directions are opposite, I'll ignore it from here.) The time to move to the edge is t=(6.0m)/(0.081m/s) = 64.2s.
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