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A 10-g bullet moving horizontally with a speed of 2.0 km/s strikes and passes through a 4.0-kg block moving with a speed of 4.2

SVEN [57.7K]

Answer:

$K=512J$

Explanation:

Since the surface is frictionless, momentum will be conserved. If the bullet of mass $m_1$ has an initial velocity $v_{1i}$ and a final velocity $v_{1f}$ and the block of mass $m_2$ has an initial velocity $v_{2i}$ and a final velocity $v_{2f}$ then the initial and final momentum of the system will be:

$p_i=m_1v_{1i}+m_2v_{2i}$

$p_f=m_1v_{1f}+m_2v_{2f}$

Since momentum is conserved, $p_i=p_f$ , which means:

$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$

We know that the block is brought to rest by the collision, which means $v_{2f}=0m/s$ and leaves us with:

$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}$

which is the same as:

$v_{1f}=\frac{m_1v_{1i}+m_2v_{2i}}{m_1}$

Considering the direction the bullet moves initially as the positive one, and writing in S.I., this gives us:

$v_{1f}=\frac{(0.01kg)(2000m/s)+(4kg)(-4.2m/s)}{0.01kg}=320m/s$

So kinetic energy of the bullet as it emerges from the block will be:

$K=\frac{mv^2}{2}=\frac{(0.01kg)(320m/s)^2}{2}=512J$

6 0

4 years ago

In part one of this experiment, a 0.20 kg mass hangs vertically from a spring and an elongation below the support point of the s

statuscvo [17]

To solve this problem it is necessary to apply the concepts related to Hooke's Law as well as Newton's second law.

By definition we know that Newton's second law is defined as

$F = ma$

m = mass

a = Acceleration

By Hooke's law force is described as

$F = k\Delta x$

Here,

k = Gravitational constant

x = Displacement

To develop this problem it is necessary to consider the two cases that give us concerning the elongation of the body.

The force to keep in balance must be preserved, so the force by the weight stipulated in Newton's second law and the force by Hooke's elongation are equal, so

$k\Delta x = mg$