ew History Bookmarks People Window Help Mar MasteringPhysics: PY211 Spr2016 HW 6
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ew History Bookmarks People Window Help Mar MasteringPhysics: PY211 Spr2016 HW 6 https://session.masteringphysics.com/myctjitemview?assignmentProbleml httpsll tD-608559588offset prev PY 211 General Physics Spring 201 Signed in as Raghav Mahm Problem 8.64 Problem 8.64 Part A A steel ball with mass 40.0 g is dropped from a height of 2.09 m onto a horizontal steel slab. The ball rebounds to a height of 1.62 m Calculate the impulse delivered to bal dng ipad N-s Part B balang bal-incontact wn he slab forab™ of240 ms.nd he awige bo.on My Answers veip 1: 20 nExplanation / Answer
F*t=m*delta-v (impulse-momentum)
if ball dropped (from rest), the vf can be calculated when it hits the ground and so can the vi when it travels back up.
vf2=vi2+2gh
vf=sqrt(2gh)
vf=sqrt(2*9.8*2.09m) = sqrt(40.964) = 6.40 m/s
also,
if the ball reaches a new max height of 1.62 m, vf=0 at max height and vi can be calculated.
vf2=vi2-2gh
2gh = vi2
vi=sqrt(2gh)
vi=sqrt(2*9.8*1.62) = 5.635 m/s
the two velocities are known, so the impulse can be calculated
impulse = F*t = m*delta-v
m*delta-v = .04 kg * (5.635 m/s - (- 6.40 m/s))
impulse = 0.4814 kg m/s
for avg force
F*t = 0.4814 kg m/s
t=.0024 s
F = 0.4814 kg m/s / .0024 s
F = 200.58 N
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