A bullet of mass m= 0.0320 kg is fired along an incline and imbeds itself quickl

Question

A bullet of mass m= 0.0320 kg is fired along an incline and imbeds itself quickly into a block of wood of mass M= 1.55 kg. The block and bullet then slide up the incline, assumed frictionless, and rise a height H= 1.65 m before stopping. Calculate the speed of the bullet just before it hits the wood. Note. The block is kept from sliding down the incline initially by as small peg.

A Princeton prof. (mass = 61.0 kg), surprised by the large stopping force he calculates for jumping flat footed from a height of 0.11 m, decides to try the experiment. Calculate he deceleration (in g's) if he stops in a distance of 0.65 cm. (Do not try this. You could easily break an ankle!)

Explanation / Answer

1)
After the collision the bullet and the block moved with a common velocity U.

from energy conservation: 1/2(m+M)U^2=(m+M)gH

U=sqrt(2gH)=sqrt(2*9.8*1.65)

U=5.6868m/sec

let the initial velocity of the bullet be V, so from conservation of momentum:

m*V=(m+M)*U

V= (m+M)*U/m

V=(0.032+1.55)*5.6868/0.032

V=281.141m/sec

2)
He falls 0.11 m and stops in .0065 m. His deceleration in G's is 0.11/0.0065 = 16.92 G (his experienced acceleration will be 1 G greater).

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