Physics 溫習題目 5

1. Robert fires a bullet of mass 0.05 kg at a block of mass 1 kg on a smooth table. The bullet takes 0.2 s to be totally embedded and to stop inside the block. After the collision, the block, together with the bullet move at a speed of 5 m/s.

a) Find the speed of the bullet just before it hits the block.

b) Find the average force acting on the block by the bullet when the bullet embeds into the block.

2. A rescue cushion allows only one person to drop on it each time. When inflated, it is 2.75 m high and can save people who drop from 30 m above the ground. Assume a person of 50 kg drops on the cushion from 30 m above the ground and is brought to a stop after time t.

a) What is the velocity of the person just before he arrives at the surface of the cushion?

b) What is the impulse acting on the person?

c) Explain briefly how the cushion saves people that fall from a height.

1 個解答

  • 天同
    Lv 7
    6 年前

    1(a) Let u be the initial speed of the bullet.

    By conservation of momentum,

    0.05u = (0.05+1) x 5

    i.e. u = 105 m/s

    (b) Use impulse = change of momentum of the block

    F(0.2) = 1 x 5 where F is the force acts on the block

    F = 25 N

    2. (a) Fall in height of the person = (30 - 2.75) m = 27.25 m

    Use equation: v^2 = u^2 + 2as

    hence, v^2 = 2g.(27.25) (m/s)^2

    where g is the acceleration due to gravity, taken to be 10 m/s^2

    hence, v = 23.35 m/s

    (b) Change of momentum of the person

    = 50 x 23.35 kg.m/s = 1167 kg.m/s

    Hence, impulse = 1167 N.s

    (c) The cushion lengthens the time of impact, thus decreasing the impact force.