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nekit [7.7K]
3 years ago
11

A 70-kg ice skater is standing still on the ice, when a friend tosses him a 10-kg ball that is traveling at +8 m/s. If he catche

s it, what would be the velocity of the person holding the ball? What type of collision is this?
Physics
1 answer:
Misha Larkins [42]3 years ago
7 0

Answer:+1.25 m/s

Explanation:

Given

mass of ice skater M=70 kg

mass of ball m=10 kg

the initial velocity of the ball  u_1=+8\ m/s

Conserving linear momentum

M\times0+m\timesu_1=(M+m)v\quad \quad [v=\text{combined velocity of skater and ball}]

v=\dfrac{10\times10}{80}=+1.25\ m/s

Therefore the velocity of the person holding the ball is 1.25 m/s

This collision represents the perfectly inelastic collision where particles stick together after the collision.

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I drop an egg from a certain distance and it takes 3.74 seconds to reach the ground. How high up was the egg?
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Answer:

68.5 meters

Explanation:

Given:

v₀ = 0 m/s

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t = 3.74 s

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A 1.05 kg block slides with a speed of 0.865 m/s on a frictionless horizontal surface until it encounters a spring with a force
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Answer:

a) U = 0 J    

k = 0.393 J

E = 0.393 J

b) U = 0.0229J

k = 0.370 J

E = 0.393 J

c) U = 0.0914 J

k = 0.302 J

E = 0.393 J

d) U = 0.206 J

k = 0.187 J

E = 0.393 J

e) U = 0.366 J

k = 0.027 J

E = 0.393 J

Explanation:

Hi there!

The equations of kinetic energy and elastic potential energy are as follows:

k = 1/2 · m · v²

U = 1/2 · ks · x²

Where:

m = mass of the block.

v = velocity.

ks = spring constant.

x = displacement of the string.

a) When the spring is not compressed, the spring potential energy will be zero:

U = 1/2 · ks · x²

U = 1/2 · 457 N/m · (0 cm)²

U = 0 J

The kinetic energy of the block will be:

k = 1/2 · m · v²

k = 1/2 · 1.05 kg · (0.865 m/s)²

k = 0.393 J

The mechanical energy will be:

E = k + U = 0.393 J + 0 J = 0.393 J

This energy will be conserved, i.e., it will remain constant because there is no work done by friction nor by any other dissipative force (like air resistance). This means that the kinetic energy will be converted only into spring potential energy (there is no thermal energy due to friction, for example).

b) The spring potential energy will be:

U = 1/2 · 457 N/m · (0.01 m)²

U = 0.0229 J

Since the mechanical energy has to remain constant, we can use the equation of mechanical energy to obtain the kinetic energy:

E = k + U

0.393 J = k + 0.0229 J

0.393 J - 0.0229 J = k

k = 0.370 J

c) The procedure is now the same. Let´s calculate the spring potential energy with x = 0.02 m.

U = 1/2 · 457 N/m · (0.02 m)²

U = 0.0914 J

Using the equation of mechanical energy:

E = k + U

0.393 J = k + 0.0914 J

k = 0.393 J - 0.0914 J = 0.302 J

d) U = 1/2 · 457 N/m · (0.03 m)²

U = 0.206 J

E = 0.393 J

k = E - U = 0.393 J - 0.206 J

k = 0.187 J

e) U = 1/2 · 457 N/m · (0.04 m)²

U = 0.366 J

E = 0.393 J

k = E - U = 0.393 J - 0.366 J = 0.027 J.

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