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A 0.145 kg block is released from rest from
the top of a 51.7◦ frictionless incline. When
it has fallen a vertical distance of 1.3 m, a
0.00794 kg bullet is fired into the block along
a path parallel to the slope of the incline, and
momentarily brings the block to rest, stopping
in the block.
The acceleration of gravity is 9.8 m/s2 .

a) Find the speed of the bullet just before

Answer in units of m/s.

b) What bullet speed is needed to send the block
up the incline to its initial position?

Answer in units of m/s.

  • physics - ,

    First calculate the speed V of the sliding block before it is hit by the bullet. You can use conservation of energy for that, since the incline is frictionless.

    (1/2) M V^2 = M g * 1.3 m
    V = sqrt(2*9.8*1.3) = 5.05 m/s

    The bullet must have an equal and opposite momentum before impact, since the momentum right after impact is zero.
    m v = M V
    where m is the bullet mass and v is the bullet velocity.

    Solve for v.

    AFTER impact, gravity will cause the bullet and block to resume falling together, but you are only worried about the instant after impact, before gravity provides an impulse to increase the momentum again.

  • physics (part b) - ,

    I did not notice your part b. That is a harder question.

    Use energy conservation to predict the velocity V' (after impact) needed to bring the block and bullet together back to the top of the incline.

    (1/2)(M+m)V'^2 = (M+m)g*1.4
    V' = 5.05 m/s (same as V in part a, but in the opposite drection)

    Now apply conversation of momentum to the impact event, and require that V' (UP the incline) be the final velocity.

    -M V + m v = (M +m) V
    v = (2M/m +1) V

  • physics - ,

    for part a:

    v= (mass of block/mass of bullet)(sqrt2gh)

    v = (0.145/0.00794) x (sqrt 2x9.8x1.3)

    v = 18.262 x 5.0478

    v= do the calculation

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