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emmainna [20.7K]

3 years ago

5

A rifle bullet with mass ma = 8.00 g strikes and embeds itself in a block with mass mb = 0.992 kg that rests on a frictionless,

horizontal surface and is attached to a coil spring. The impact compresses the spring Δx = 15.0 cm. Calibration of the spring shows that a force of 0.750 N is required to compress the spring 0.250 cm.
(a) Find the magnitude of the velocity of the block just after impact.
(b) What was the original speed of the bullet?

1 answer:

Doss [256]3 years ago

6 0

Answer:

the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.

Explanation:

a) Kinetic energy of block = potential energy in spring

½ mv² = ½ kx²

Here m stands for combined mass (block + bullet),

which is just 1 kg. Spring constant k is unknown, but you can find it from given data:

k = 0.75 N / 0.25 cm

= 3 N/cm, or 300 N/m.

From the energy equation above, solve for v,

v = v √(k/m)

= 0.15 √(300/1)

= 2.598 m/s.

b) Momentum before impact = momentum after impact.

Since m = 1 kg,

v = 2.598 m/s,

p = 2.598 kg m/s.

This is the same momentum carried by bullet as it strikes the block. Therefore, if u is bullet speed,

u = 2.598 kg m/s / 8 × 10⁻³ kg

= 324.76 m/s.

Hence, the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.

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