Ask question

Ask question

All categories

2 years ago

13

Object A of mass M is moving east at speed v. It collides with object B of mass 2M that was initially at rest. The motion of the

objects before and after the collision is along the same line. After the collision, object A is moving west at a speed of v/3. What is the speed of object B immediately after the collision?

1 answer:

Firlakuza [10]2 years ago

3 0

Answer:

$v_B=\frac{v}{3}$

Explanation:

Given that:

mass of object A, $m_A=M$

mass of object B, $m_B=2M$

speed of object A, $v_A=v$

So, according to the conservation of momentum, the momentum before collision is equal to the momentum after conservation.

$m_A.v+m_B\times 0=m_A\times v_A +m_B\times v_B$

$M\times v+0 = M\times \frac{v}{3}+2M\times v_B$

$2M\times v_B= \frac{2M\times v}{3}$

$v_B=\frac{v}{3}$

You might be interested in

While at a party, you pull up a sound intensity level app on your phone (everyone does stuff like that, right?), and it reads 83

allochka39001 [22]

To solve this problem it is necessary to apply the concepts related to Sound Intensity. The unit most used in the logarithmic scale is the decibel and mathematically this is expressed as

$\beta_{dB} = 10log_{10}\frac{I}{I_0}$

Where,

$\beta_{dB}$ = Sound intensity level in decibels

I = Acoustic intensity on the linear scale

$I_0 =$ Hearing threshold

According to the values, the total intensity is 32 times the linear intensity and the value in decibels is 83dB

So:

$10log_{10}(\frac{32I}{I_0}) = 83$

$10log(\frac{I}{I_0})+10log(32) = 83$

$10log(\frac{I}{I_0})= 83-10log(32)$

$10log(\frac{I}{I_0})= 67.948dB$

Therefore the sound intensity due to one person is 67.948dB

8 0

2 years ago

Four wires meet at a junction. In two of the wires, currents and enter the junction. In one of the wires, current leaves the jun

Vera_Pavlovna [14]

Answer:

Explanation:

Kirchoff's current law: It states that the total current entering at a junction is equal to the total current leaving at the junction. It isbased on conservation of charge.

As two current i1 and i2 is entering at the junction and i3 is leaving, so the current in forth wire is

i4 = i1 + i2 - i3

4 0

2 years ago

24 A person stands at the side of a straight railway track. A train moves towards the person and emits sound from its whistle. T

Andrei [34K]

Bffndjdjdjdkdjcfndjndkfkfk

6 0

2 years ago

Assuming the ball's initial velocity was 51 ∘ above the horizontal and ignoring air resistance, what did the initial speed of th

Umnica [9.8K]

horizontal distance of home run is 400 ft = 122 m

height of the home run is 3 ft = 0.9 m

now the angle of the hit is 51 degree

now we have equation of trajectory of the motion

$x = vcos\theta * t$

$y = v sin\theta * t - \frac{1}{2} gt^2$

solving above two equations we have

$y = xtan\theta - \frac{gx^2}{2v^2cos^2\theta}$

now here we will plug in all data

$0.9 = 122 tan51 - \frac{9.8 * 122^2}{2*v^2 * cos^251}$

$0.9 = 150.65 - \frac{184150.2}{v^2}$

$\frac{184150.2}{v^2} = 149.75$

$v = 35.1 m/s$

<em>so the ball was hit with speed 35.1 m/s from the ground</em>

8 0

2 years ago

a car travelling at 50km/h from rest covers a distance of 10km in 40minutes. Calculate the acceleration

Margarita [4]

Answer:

$9.67\cdot 10^{-3}m/s^2$

Explanation:

We can solve the problem by using the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

For the car in this problem:

u = 0 (it starts from rest)

$v=50 km/h \cdot \frac{1000}{3600}=13.9 m/s$ is the final velocity

s = 10 km = 10 000 m is the displacement

Solving for a, we find:

$a=\frac{v^2-u^2}{2s}=\frac{13.9^2}{2(10000)}=9.67\cdot 10^{-3}m/s^2$

7 0

2 years ago