Physics

An 8.00 g bullet is fired into a 250 g block that is initially at rest at the edge of a table of 1.00 m height (Fig. P6.26). The bullet remains in the block, and after the impact the block lands 2.00 m from the bottom of the table. Determine the initial speed of the bullet.

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  1. So the block fell through a height of 1 m in time t, where
    1.0=(1/2)gt²
    or
    t=√(2/9.81)
    =0.452 s

    During this time, the block has travelled horizontally 2m, or
    horiz. velocity
    =2/0.452
    =4.43 m/s

    This velocity, v, is the velocity of the block after impact, given by the equation of momentum before and after impact, namely
    m1u1+m2u2=(m1+m2)v
    or
    v=(m1u1+m2u2)/(m1+m2)
    where
    m1=8g
    m2=250g
    u1= to be determined
    u2=0 (block)
    Solve for u1 (velocity of bullet)

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  2. do the end first
    mass m (does not matter what m is) lands 2 meters from the table

    how long did it fall?
    h = 0 at end
    0 = 1 - (1/2)(9.8) t^2
    2 = 9.8 t^2
    t = .45 seconds to fall to floor off table = time in air
    distance = speed*time
    2 = speed* .45
    speed = 4.4 m/s horizontal
    .008 v = .258 (4.4)
    v = 141.9 m/s

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  3. 48.Consider a frictionless track as shown in Figure. A block of mass m1 = 5.00 kg is released from A. It makes a head-on elastic collision at B with a block of mass m2 =10.0 kg that is initially at rest. Calculate the maximum height to which m1 rises after the collision.

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  4. An 8.00 g bullet is fired into a 250 g block that is initially at rest at the edge of a table of 1.00 m height (Fig. P6.26). The bullet remains in the block, and after the i

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