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Posted by on Sunday, December 23, 2012 at 7:40pm.

A(n) 15 g bullet is fired into a(n) 116 g block
of wood at rest on a horizontal surface and
stays inside. After impact, the block slides
15 m before coming to rest.
The acceleration of gravity is 9.8 m/s
2
.
If the coefficient of friction between the
surface and the block is 0.7, find the speed of
the bullet before impact.
Answer in units of m/s

  • Physics - , Monday, December 24, 2012 at 7:26pm

    The kinetic energy of the (bullet+block) after impact equals the work done against friction at the point where it stops. Solve for the velocity right after impact, V'. Then use conservatiuon of momentum to solve for the bullet velocity, V.

    (1/2)(m+M) V'^2 = (m+M)*Mu*g*X
    V' = 14.35 m/s

    m*V = (m+M)*V'

  • Physics - , Sunday, November 2, 2014 at 1:21am

    What does the M in Mu stand for in the equation? Momentum?

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