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A bullet of mass 10gm,moving horizantally,stries and embedded itself in a box of mass 1Kg suspended from a light string as shown in the height 0.45,then what is the speed of the bullet before collision?

Sagot :

MathPhys

Answer:

300 m/s

Explanation:

The momentum of the bullet and box is conserved before and after the collision, where momentum (p) is equal to mass (m) times velocity (v). After the collision, energy is conserved as the initial kinetic energy (KE) is converted to gravitational potential energy (PE). Kinetic energy is half the mass (m) times the square of the speed (v), and gravitational potential energy is equal to weight (mg) times height (h).

Defining variables:

  • m is the mass of the bullet
  • M is the mass of the box
  • u is the initial speed of the bullet
  • v is the speed of the bullet/box after the collision
  • h is the height of the bullet/box

Starting with conservation of energy:

KE = PE

½ (M + m) v² = (M + m) gh

½ v² = gh

v² = 2gh

v² = 2 (10 m/s²) (0.45 m)

v = 3.0 m/s

Next, momentum is conserved during the collision.

p₀ = p

mu = (M + m) v

u = (M + m) v / m

u = (1 kg + 0.01 kg) (3.0 m/s) / (0.01 kg)

u = 303 m/s

Rounded to one significant figure, the initial speed of the bullet is 300 m/s.

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