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3 years ago

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a piece of clay with mass m is traveling with an unknown velocity in the positive x-direction, when it collides and sticks to a

block of mass 3m, moving with velocity v=-3m/s. after the collision, the block and clay are stuck together and traveling with a velocity of 1 m/s in the negative x-direction. what was the original velocity of the clay mass?

1 answer:

joja [24]3 years ago

7 0

Answer:

13 m in the + x direction.

Explanation:

Givens

m1 = m

m2 = 3m

v1= ? = x

v2 = - 3 m/s

v3 = v4 = - 1 m/s

Equation

Let x be the speed of the lighter object before collision.

Momentum before collision = momentum after collision

Solution

mx - 3m * 3m/s = (m1 + m2)*(-1)

m(x - 3*3m/s) = (m + 3m)(-1)

m(x - 9) = - 4m Cancel the ms

x - 9 = 4 Add 9 to both sides

x = 9 + 4

x = 13 m/s

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3 years ago

Calculate the separation between the two lowest levels for an O2 molecule in a one-dimensional container of length 5.0 cm. At wh

MariettaO [177]

Answer:

The separation between the two lowest levels = $1.24 * 10^{-39}J$

The values of n where the energy of molecule reaches 1/2 kT at 300K = $2.2 * 10^{9}$

The separation at this level = 1.8 * $10^{-30}$ J

Explanation:

Knowing the formula

En = $\frac{n^{2} h^{2} }{8 mL^{2} }$

Mass of oxygen molecule

m (O2) = 32 amu * $\frac{1.6605 * 10^{-27 kg} }{1 amu}$

So the energy diference between the two lowest levels:

E2 - E1 = $\frac{3h^{2} }{8mL^{2} }$

E2 - E1 = $\frac{3 * (6.626 * 10^{-34} Js)^{2} } {8 * 32 amu * (\frac{1.6605 * 10^{-27 kg} }{1 amu})* (5*10^{-2})^{2} } = 1.24 * 10^{-39}J$

Now we should find n where the energy of molecule reaches 1/2 kT

En = $\frac{n^{2} h^{2} }{8 mL^{2} }$ = $\frac{1}{2}kT$

$\frac{h^{2} }{8 mL^{2} }$ = $4.13 * 10^{-14}J$

$n^{2} * (4.13 * 10^{-14}J) = \frac{1}{2} (1.38 * 10^{-23}JK^{-1}) * 300K$

$n = 2.2 * 10^{9}$

by the end is necessary to calculate the separation of the level

En - En-1 = $(n^{2} - (n - 1)^{2}) * \frac{h^{2} }{8 mL^{2} }$

= 1.8 * $10^{-30}$ J

4 0

3 years ago

You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics

olga nikolaevna [1]

Answer:112.82 m/s

Explanation:

Given

range of arrow=68 m

$Angle=3^{\circ}$

as the arrow travels it acquire a vertical velocity $v_y$

$v_y=u+at$

$v_y=0+9.81\times t$ -------1

Range is given by

R=ut

where u=initial velocity

$68=u\times t$

$t=\frac{68}{u}$

substitute the value of t in eqn 1

$v_y=9.81\times \frac{68}{u}$

$v_y\times u=9.81\times 68=667.08$ --------2

and $tan(3)=\frac{v_y}{u}$

$v_y=utan(3)=0.0524u$

substitute it in 2

$0.0524 u^2=667.08$

$u^2=12,728.644$

u=112.82 m/s

6 0

3 years ago

To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T

Fantom [35]

Answer:

$V_1= 3.4*10^7m/s$

Explanation:

From the question we are told that

Nucleus diameter $d=5.50-fm$

a 12C nucleus

Required kinetic energy $K=2.30 MeV$

Generally initial speed of proton must be determined,applying the law of conservation of energy we have

$K_2 +U_2=K_1+U_1$

where

$K_1$ =initial kinetic energy

$K_2$ =final kinetic energy

$U_1$ =initial electric potential

$U_2$ =final electric potential

mathematically

$U_2 = \frac{Kq_pq_c}{r_2}$

where

$r_f$ =distance b/w charges

$q_c$ =nucleus charge $=6(1.6*10^-^1^9C)$

$K$ =constant

$q_p$ =proton charge

Generally kinetic energy is know as

$K=\frac{1}{2} mv^2$

Therefore

$U_2 = \frac{Kq_pq_c}{r_2} + K_2=\frac{1}{2} mv_1^2 +U_1$

Generally equation for radius is $d/2$

Mathematically solving for radius of nucleus

$R=(\frac{5.50}{2}) (\frac{1*10^-^1^5m}{1fm})$

$R=2.75*10^-^1^5m$

Generally we can easily solving mathematically substitute into v_1

$q_p$ $=6(1.6*10^-^1^9C)$

$K_1=9.0*10^9 N-m^2/C^2$

$U_1= 0$

$R=2.75*10^-^1^5m$

$K=2.30 MeV$

$m= 1.67*10^-^2^7kg$

$V_1= (\frac{2}{1.67*10^-^2^7kg})^1^/^2 (\frac{(9.0*10^9 N-m^2/C^2)*(6(1.6*10^-^1^9C)(1.6*10^-^1^9C)}{2.75*10^-^1^5m+2.30 MeV(\frac{1.6*10^-^1^3 J}{1 MeV}) }$

$V_1= 3.4*10^7m/s$

Therefore the proton must be fired out with a speed of $V_1= 3.4*10^7m/s$

8 0

3 years ago