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bonufazy [111]
3 years ago
10

A 5 kg ball rolling at 1.0 m/s hits a 15 kg ball at rest. The balls stick together after the collision What is the

Physics
1 answer:
Reika [66]3 years ago
5 0

Answer:

0.25m/s

Explanation:

Given parameters

m₁  = 5kg

v₁ = 1.0m/s

m₂ = 15kg

v₂ = 0m/s

Unknown:

velocity after collision = ?

Solution:

Momentum before collision and after collision will be the same. For inelastic collision;

    m₁v₁ + m₂v₂  = v(m₁ + m₂)

Insert parameters and solve for v;

   5 x 1  + 15 x 0 = v (5 + 15 )

          5  = 20v

          v = \frac{5}{20}   = 0.25m/s

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O-penness.

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2 years ago
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A large fake cookie sliding on a horizontal surface is attached to one end of a horizontal spring with spring constant k = 440 N
irinina [24]

Answer:

a) 0.275 m b) 13.6 J

Explanation:

In absence of friction, the energy is exchanged between the spring (potential energy) and the cookie (kinetic energy), so at any point, the sum of both energies must be the same:

E = ½ kx2 + ½ mv2

If we take as initial state, the instant when the cookie is passing through the spring’s equilibrium position, all the energy is kinetic, and we know that is equal to 20.0 J.

After sliding to the right, while is being acted on by a friction force, it came momentarily at rest. At this point, the initial kinetic energy, has become potential elastic energy, in part, and in thermal energy also, represented by the work done by the friction force.

So, for this state, we can say the following:

Ki = Uf + Eth = ½* k*d2 + Ff*d

20.0J = ½ *440 N/m* d2 + 11.0 *d, where d is the compressed length of the spring, which is equal to the distance travelled by the cookie before coming momentarily at rest.

We have a quadratic equation, that, after simplifying terms, can be solved as follows, applying the quadratic formula:

d = -0.05/2 +/- √0.090625 = -0.025 +/- 0.3 = 0.275 m (we take the positive root)

b) If we take as our new initial status the moment at which the spring is compressed, and the cookie is at rest, all the energy is potential:

E = Ui = 1/2 k d²

In this case, d is the same value that we got in a), i.e., 0.275 m (as the distance travelled by the cookie after going through the equilibrium point is the same length that the spring have been compressed).

E= 1/2 440 N/m . (0.275)m² = 16.6 J

When the cookie passes again through the equilibrium position, the energy will be in part kinetic, and in part, it will have become thermal energy again.

So, we can write the following equation:

Kf = Ui - Ff.d = 16.6 J - 11.0 (0.275) m = 16.6 J - 3.03 J = 13.6 J

3 0
3 years ago
Through what voltage must an αα-particle, with its charge of +2e+2e, be accelerated so that it has just enough energy to reach a
mezya [45]

Answer:5101.35v

Explanation:

Radius of gold nucleus=7.3×10-15m and a charge of +79e

Q= 79e

e=1.6×10^-19

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The nucleus is considered as the point charge where the potential energy between the charges are

U = 1/(4×3.142×Eo)×(qQ)/r

Where r is distance between the charges and the nucleus

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V=U/q

U= 1/(4×3.142×Eo)×Q/r

V= 1/(4×3.142×Eo)×Q/(r+d)

9.0×10^9 ×(79×10^-19)/(7.3×10^-15)+(1.5×10^-14)

V= 9.0×10^9 ×(1.264×10^-17)/(2.23×10^-14)

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Answer:

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