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denpristay [2]
2 years ago
9

A 30-g bullet traveling horizontally at 300 m/s strikes a 1.0-kg block which is attached to a horizontal spring with a force con

stant of 2000 N/m and rests on a frictionless horizontal surface. The spring is on the far side of the block and aligned with the direction of travel of the bullet. The bullet becomes embedded in the block and, as a result of the impact, and the block slides against the spring. How far is the spring compressed before it reverses its direction of travel
Physics
1 answer:
snow_lady [41]2 years ago
7 0

Answer:

Explanation:

Using law of conservation of momentum during collision , velocity after collision can be found .

common velocity V = m v / ( m + M )

= .030 x 300 / ( .030 + 1 )

V= 8.7378 m /s

The kinetic energy of the bullet and block will be stored as elastic energy

1/2 ( M + m ) V² = 1/2 k A² where k is force constant and A is maximum compression in the spring .

( M + m ) V² =  k A²

1.030 x 8.7378² = 2000 x A²

A² = .0393

A= .1982 m .

= 19.82 cm

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Answer:

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A boy weighs 40 kilograms. He runs at a velocity of 4 meters per second north. Which is his momentum?
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4 0
3 years ago
A child kicks a ball horizontally with a speed of 4.8 m/s off a deck 3.5 m off the ground. How far, in meters, from the deck doe
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Answer:

2.605m

Explanation:

Using the formula for calculating Range (distance travelled in horizontal direction)

Range R = U√2H/g

U is the speed = 4.8m/s

H is the maximum height = ?

g is the acc due to gravity = 9.8m/s²

R = 3.5m

Substitute into the formula and get H

3.5 = 4.8√2H/9.8

3.5/4.8 = √2H/9.8

0.7292 = √2H/9.8

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0.7292² = 2H/9.8

2H = 0.7292² * 9.8

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