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3 years ago

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A 4.0-kilogram ball moving at 8.0 m/s to the right collides with a 1.0-kilogram ball at rest. After the collision, the 4.0-kilog

ram ball moves at 4.8 m/s to the right. What is the velocity of the 1-kilogram ball?

2 answers:

ELEN [110]3 years ago

7 0

<u>Answer:</u> The velocity of ball having less mass is 12.8 m/s

<u>Explanation:</u>

To calculate the velocity of the ball having less mass after the collision, we use the equation of law of conservation of momentum, which is:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

$m_1$ = mass of ball 1 = 4.0 kg

$u_1$ = Initial velocity of ball 1 = 8.0 m/s

$v_1$ = Final velocity of ball 1 = 4.8 m/s

$m_2$ = mass of ball 2 = 1.0 kg

$u_2$ = Initial velocity of ball 2 = 0 m/s

$v_2$ = Final velocity of ball 2 = ?

Putting values in above equation, we get:

$(4.0\times 8.0)+(1.0\times 0)=(4.0\times 4.8)+(1.0\times v_2)\\\\v_2=\frac{32-19.2}{1}=12.8m/s$

Hence, the velocity of ball having less mass is 12.8 m/s

Illusion [34]3 years ago

3 0

Its about momentum. Momentum (p)=mass(m)xvelocity(v)
So for the first ball P=4x8=32kgm/s
For the second the momentum is zero as it is still.
So overall momentum its 32kgm/s
Momentum has to be conserved
After the collision the momentum of the 4kg ball is 4x4.8=19.2kgm/s
As momentum is conserved 32-19.2=12.8kgm/s remaining
So rearrange for velocity so v=p/m=12.8/1=12.8m/s for the 1kg ball

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When a body of mass 0.25 kg is attached to a vertical massless spring, it is extended 5.0 cm from its unstretched length of 4.0

lora16 [44]

Answer:

$d=0.165m$

Explanation:

Given

$m=0.25kg$ , $x_{1}=5cm*\frac{1m}{100cm}=0.05m$ , $x_{2}=4cm*\frac{1m}{100cm}=0.04m$ , $v=2\frac{rev}{s}$

The tension of the spring is

$F_{k}=K*x_{1}=m*g$

$K=\frac{m*g}{x_{1}}$

$K=\frac{0.25kg*9.8m/s^2}{0.05m}=49N/m$

The force in the spring is equal to centripetal force so

$F_{c}=\frac{m*v^2}{r}$

$v=w*r=2\pi*r$

But Fc is also

Fc=KxΔr

$F_{c}=K*(r-x_{2})$

Replacing

$m*4\pi^2*r=K*(r-x_{2})$

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3 0

3 years ago

Two cars are moving with velocities 70km/hr and 50km/hr in east and west direction respectively.

viva [34]

Answer:

70 + 50 = 120 km/hr

Explanation:

The driver of either car would see the other car approaching or departing at 120 kph

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3 years ago

You are investigating an elevator accident which happened in a tall building. An elevator in this building is attached to a stro

Advocard [28]

Answer:

a) F = 2250 Ib

b) F = 550 Ib

c) new max force ( F newmax ) = 2850 Ib

Explanation:

A) The force the wall of the elevator shaft exert on the motor if the elevator starts from rest and goes up

max capacity of elevator = 24000 Ibs

counterweight = 1000 Ibs

To calculate the force (F) :

we first calculate the Tension using this relationship

Counterweight (1000) - T = ( 1000 / g ) ( g/4 )

Hence T = 750 Ib

next determine F

750 + F - 2400 = 2400 / 4

hence F = 2250 Ib

B ) calculate Tension first

T - 1000 = ( 1000/g ) ( g/4)

T = 1250 Ib

F = 2400 -1250 - 2400/ 4

F = 550 Ib

C ) determine design limit

Max = 2400 * 1.2 = 2880 Ib

750 + new force - 2880 = 2880 / 4

new max force ( F newmax ) = 2850 Ib

8 0

4 years ago

A dog pushes against the front door for 3 seconds with a force of 88 N. What

kirill [66]

264Ns

Explanation:

Given parameters:

Time of the push = 3s

Force of push = 88N

Unknown:

Impulse = ?

Solution:

Impulse is defined as the change in momentum of a body when force acts on it.

Impulse = Force x time

Inputting the parameters:

Impulse = 88 x 3 = 264Ns

Learn more:

Momentum brainly.com/question/9484203

#learnwithBrainly

5 0

3 years ago

Water (density = 1x10^3 kg/m^3) flows at 15.5 m/s through a pipe with radius 0.040 m. The pipe goes up to the second floor of th

RUDIKE [14]

Answer:

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

Explanation:

By assuming that fluid is incompressible and there are no heat and work interaction through the line of current corresponding to the pipe, we can calculate the speed of the water floor in the pipe on the second floor by Bernoulli's Principle, whose model is:

$P_{1} + \frac{\rho\cdot v_{1}^{2}}{2}+\rho\cdot g\cdot z_{1} = P_{2} + \frac{\rho\cdot v_{2}^{2}}{2}+\rho\cdot g\cdot z_{2}$ (1)

Where:

$P_{1}$ , $P_{2}$ - Pressures of the water on the first and second floors, measured in pascals.

$\rho$ - Density of water, measured in kilograms per cubic meter.

$v_{1}$ , $v_{2}$ - Speed of the water on the first and second floors, measured in meters per second.

$z_{1}$ , $z_{2}$ - Heights of the water on the first and second floors, measured in meters.

Now we clear the final speed of the water flow:

$\frac{\rho\cdot v_{2}^{2}}{2} = P_{1}-P_{2}+\rho \cdot \left[\frac{v_{1}^{2}}{2}+g\cdot (z_{1}-z_{2}) \right]$

$\rho\cdot v_{2}^{2} = 2\cdot (P_{1}-P_{2})+\rho\cdot [v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})]$

$v_{2}^{2}= \frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})$

$v_{2} = \sqrt{\frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2}) }$ (2)

If we know that $P_{1}-P_{2} = 0\,Pa$ , $\rho=1000\,\frac{kg}{m^{3}}$ , $v_{1} = 15.5\,\frac{m}{s}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $z_{1}-z_{2} = -3.5\,m$ , then the speed of the water flow in the pipe on the second floor is:

$v_{2}=\sqrt{\left(15.5\,\frac{m}{s} \right)^{2}+2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (-3.5\,m)}$

$v_{2} \approx 13.100\,\frac{m}{s}$

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

4 0

3 years ago