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krok68 [10]
3 years ago
5

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

Physics
1 answer:
Usimov [2.4K]3 years ago
8 0

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

u=6.647m/s

v=1.534m/s

<h3> Velocity of the bullet and the block</h3>

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

v=\sqrt{2*9.8*0.12}

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

brainly.com/question/15959704

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