Physics, 17.03.2020 16:51 andrewjammo

A 9.00-gg bullet is fired horizontally into a 1.20-kgkg wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.20. The bullet remains embedded in the block, which is observed to slide 0.360 mm along the surface before stopping. What was the initial speed of the bullet?

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Answer from: faithclark0

$u=504.6358\ m.s^{-1}$ is the initial velocity of the bullet.

Explanation:

Given:

mass of the bullet, $m=0.009\ kg$ final velocity of the bullet, $v=0\ m.s^{-1\\}$ mass of the wooden block, $m'=1.2\ kg$ initial velocity of the block, $u'=0\ m.s^{-1}$ coefficient of friction between the block and the surface, $\mu_k=0.2$ displacement of the block on the surface due to impact, $s=0.36\ mm$

Since here there is no mention of the penetration by the bullet therefore we consider the bullet to be a point that sticks on the surface of the block as soon as it collides.

From the conservation of linear momentum:

Momentum of the system just before collision = momentum of the system just after collision

m.u+m'.u'=(m.+m').v'

$0.009\times u+0=(0.009+1.2)\times v'$ .............................(1)

where:

v'= final velocity of the combined mass of block and bullet

Now the frictional force will be due to combined mass:

$f=\mu.N$

where:

normal reaction from the surface due to weight

$f=\mu.(m+m')g$

$f=0.2\times (1.2+0.009)\times 9.8$

$f=2.36964\ N$ This is the force that brings deceleration in the combined mass.

Now the deceleration rate:

$a=\frac{f}{(m+m')}$

$a=\frac{2.36964}{1.209}$

$a=1.96\ m.s^{-2\\}$ Since the force of friction acts in the opposite direction so this acceleration will also act in the same direction.

Now using the equation of motion for the course of motion when the masses just combine after collision:

v"^2=v'^2-2a.s

(using -ve sign because the acceleration opposes the motion here)

initial velocity of the combined mass just after collision v"=

final velocity of the combined mass = 0

$0^2=v'^2-2\times 1.96\times 3.6\times 10^{-4}$

$v'=3.7566\ m.s^{-1\\}$

Now using eq. (1):

$0.009\times u+0=(0.009+1.2)\times 3.7566$

$u=504.6358\ m.s^{-1}$ is the initial velocity of the bullet.

Answer from: Quest

the assestment will probley apele with cause and effect

Answer from: Quest

stands for electromotive force and it is measured in volts. despite its name, it is not actually the force in the sense of being a "pushing" or "pulling" (not measured in newtons).an equation that relates emf, ℰ, to a battery's terminal voltage, v, current, i, and internal resistance, r, is ℰ = v + irin this case, the emf can be interpreted as a voltage—a difference in electric potential.i would not say that it would be considered an energy. if emf can be defined as the energy/work per a unit charge from a battery or other electric source, then it is not strictly energy.

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A 9.00-gg bullet is fired horizontally into a 1.20-kgkg wooden block resting on a horizontal surface...