Conservation of energy minimum speed

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firezap
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Homework Statement


Gus is at end of rope. He needs to reach roof. His friend fires cannon ball horizontally hitting Gus and get stuck in belt. Find minimum speed of cannon ball for Gus to reach roof. 50m is rope. 99kg is Gus. 1kg is ball. 45° angle in diagram


Homework Equations


conservation of energy
Et1 = Et2
1/2mv^2 = mgh
momentum
p = mv


The Attempt at a Solution


1/2(1)v^2 = 99(9.8)(50xsin45°)
1/2v^2 = 34301.75
v = 260 m/s
is this right?
 
Attachments
  • Diagram of a pendulum: a 50 m rope hangs from a fixed point with a 99 kg mass at its end, swung out to 45 degrees to a dashed position beside a rectangular platform. A 1 kg ball approaches horizontally from the left.
    mspaint2.jpg
    11 KB · Views: 492
Last edited:
conservation of energy:
1/2 (m+M) V² = (m+M)gh
V² = 2gh
conservation of momentum:
mv = (m+M)V
v = (m+M) sqrt(2gh) / m
is this correct?
 
firezap said:
conservation of energy:
1/2 (m+M) V² = (m+M)gh
V² = 2gh
conservation of momentum:
mv = (m+M)V
v = (m+M) sqrt(2gh) / m
is this correct?
Looks good. The collision of cannonball and person is treated as an inelastic collision.