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In the ballistic pendulum experiment, a bullet of mass 0.06 kg is fired
horizontally into a wooden block of mass 0.2 kg. The wooden block is
suspended from the ceiling by a long string. The collision is perfectly
inelastic and after impact the bullet and the block swing together until
the block is 0.12 m above its initial position. Find the velocity of the
bullet and the block just after impact. Round your answer to the
nearest tenth.

Respuesta :

okpalawalter8

The velocity of the bullet and the block just after impact is mathematically given as

u=6.647m/s

v=1.534m/s

Velocity of the bullet and the block

Question Parameters:

In the ballistic pendulum experiment, a bullet of mass 0.06 kg is fired

horizontally into a wooden block of mass 0.2 kg.

the block is 0.12 m above its initial position.

Generally the equation for the conservation of energy  is mathematically given as

0.5(m+n)v^2=(m+n)gh

Therefore

v^2=2gh

[tex]v=\sqrt{2*9.8*0.12}[/tex]

v=1.534m/s

Hence

mu+m*0=(m+n)v

0.060*u=(0.060+0.2)*1.534

u=6.647m/s

For more information on mass

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