Physics, 15.04.2021 15:40 jimena15

1 kg block slides down a frictionless inclined plane that makes an angle of 300 with respect to the ground. The total length of the plane is 2 m, but midway down it collides with a second block, weighing 0.5 kg. The two blocks stick together and travel as one unit the rest of the way down the ramp. What is the kinetic energy of the combined 1.5 kg block when it reaches the bottom of the plane

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Answer from: layshascott

the kinetic energy of the combined 1.5 kg block when it reaches the bottom of the plane is 10.62 J

Explanation:

Given that the data in the question;

angle of inclination with respect to the ground $\theta$ = 30°

length of plane d = 2m

m₁ = 1 kg

m₂ = 0.5 kg

now, velocity of the first block at midpoint;

$\frac{1}{2}$ mv² = mgsin $\theta$ $\frac{d}{2}$

$\frac{1}{2}$ v² = gsin $\theta$ $\frac{d}{2}$

v² = gsin $\theta$ d

v = √( gsin $\theta$ d)

g is 9.8 m/s

so we substitute

v = √( 9.8 × sin30° × 2)

v = √( 19.6 )

v = 3.13 m/s

Now, velocity just after collision of the blocks will be;

(m₁ + m₂)v₂ = m₁v

v₂ = m₁v / (m₁ + m₂)

we substitute

v₂ = (1 × 3.13) / (1 + 0.5)

v₂ = 3.13 / 1.5

v₂ = 2.0866 m/s

now, final kinetic energy will be;

KE_f = (m₁ + m₂)gsin $\theta$ $\frac{d}{2}$ + Initial Kinetic energy

KE_f = (m₁ + m₂)gsin $\theta$ $\frac{d}{2}$ + $\frac{1}{2}$ mv₂²

we substitute

KE_f = [(1 + 0.5)9.8 × sin30 × $\frac{2}{2}$ ] + [ $\frac{1}{2}$ × 1.5 × 2.0866 ]

KE_f = 7.35 + 3.2654

KE_f = 10.62 J

Therefore, the kinetic energy of the combined 1.5 kg block when it reaches the bottom of the plane is 10.62 J

Answer from: Quest

Distance^2= yellow line ^2 (see picture) = 30^2 + 49^2so distance = squareroot of (300^2 +49^2)i dont know if this is how to solve it, but i cnt think of another way

Answer from: Quest

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

$f\alpha \frac{dp}{dt}$

⇒ $f= k \frac{dp}{dt}$

here k is the proportionality constant whose value is 1.

$f=\frac{dp}{dt}$

$f= \frac{d[mv]}{dt}$

$f= m\frac{dv}{dt} +v\frac{dm}{dt}$

$f=ma+v\frac{dm}{dt}$ [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.

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