In an elastic collision, the momentum is _____, and the mechanical energy is ____

A 500 kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 30 N/m. The blo

m_a_m_a [10]

Answer:

x = 0.396 m

Explanation:

The best way to solve this problem is to divide it into two parts: one for the clash of the putty with the block and another when the system (putty + block) compresses it is spring

Data the putty has a mass m1 and velocity vo1, the block has a mass m2 . t's start using the moment to find the system speed.

Let's form a system consisting of putty and block; For this system the forces during the crash are internal and the moment is preserved. Let's write the moment before the crash

p₀ = m1 v₀₁

Moment after shock

$p_{f}$ = (m1 + m2) $v_{f}$

p₀ = $p_{f}$

m1 v₀₁ = (m1 + m2) $v_{f}$

$v_{f}$ = v₀₁ m1 / (m1 + m2)

$v_{f}$ = 4.4 600 / (600 + 500)

$v_{f}$ = 2.4 m / s

With this speed the putty + block system compresses the spring, let's use energy conservation for this second part, write the mechanical energy before and after compressing the spring

Before compressing the spring

Em₀ = K = ½ (m1 + m2) $v_{f}$ ²

After compressing the spring

$E_{mf}$ = Ke = ½ k x²

As there is no rubbing the energy is conserved

Em₀ = $E_{mf}$

½ (m1 + m2) $v_{f}$ ² = = ½ k x²

x = $v_{f}$ √ (k / (m1 + m2))

x = 2.4 √ (11/3000)

x = 0.396 m

7 0

3 years ago

A ball having a mass of 200 g is released from rest at a height of 400 mm above a very large fixed metal surface. If the ball re

AysviL [449]

Answer:

0.9

Explanation:

h = 400 mm, h' = 325 mm

Let the coefficient of restitution be e.

h' = e^2 x h

325 = e^2 x 400

e^2 = 0.8125

e = 0.9

5 0

3 years ago

To construct a solenoid, you wrap insulated wire uniformly around a plastic tube 7.1 cm in diameter and 57 cm in length. You wou

ASHA 777 [7]

Answer:

We need about 8769 meters of wire to produce a 2.6 kilogauss magnetic field.

Explanation:

Recall the formula for the magnetic field produced by a solenoid of length L. N turns, and running a current I:

$B=\mu_0\,\frac{N}{L} \,I$

So, in our case, where B = 2.6 KG = 0.26 Tesla; I is 3 amperes, and L = 0.57 m, we can find what is the number of turns needed;

$B=\mu_0\,\frac{N}{L} \,I\\0.26=4\,\pi\,10^{-7}\frac{N}{0.57} \,3\\N=\frac{0.26*0.57\,10^7}{12\,\pi} \\N=39311.27$

Therefore we need about 39312 turns of wire. Considering that each turn must have a length of $\pi\,D$ , where D is the diameter of the plastic cylindrical tube, then the total length of the wire must be: