Is the change in position of an object

Which of the following statements about the conservation of momentum is not correct? Momentum is conserved for a system of objec

Olin [163]

Answer:

wrong statement : Momentum is not conserved for a system of objects in a head-on collision.

Explanation:

In a head on collision of two objects , two equal and opposite forces are created at the point of collision . These two forces create two impulses in opposite direction which results in equal and opposite changes in momentum in each of them . Hence net change in momentum is zero. In this way momentum is conserved in head on collision of two objects.

3 0

3 years ago

A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike

Anvisha [2.4K]

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g= $9.50\times 10^{3} kg$

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

$mu+ MU=mv+ MV$

Substitute the values

$9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V$

$3.61i=2.375j+0.150V$

$3.61 i-2.375j=0.150V$

$V=\frac{1}{0.150}(3.61 i-2.375j)$

$V=24.07i-15.83j$

Magnitude of velocity of stone

= $\sqrt{(24.07)^2+(-15.83)^2}$

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

$\theta=tan^{-1}(\frac{y}{x})$

= $tan^{-1}(\frac{-15.83}{24.07})$

$\theta=tan^{-1}(-0.657)$

=33.3 degree below the horizontal

(b)

Initial kinetic energy

$K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2$

$K_i=685.9 J$

Final kinetic energy

$K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2$

= $\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2$

$K_f=359.12 J$

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

5 0

2 years ago

A batter hits a foul ball. The 0.140-kg baseball that was approaching him at 40.0 m/s leaves the bat at 30.0 m/s in a direction

lara31 [8.8K]

<h3>Answer</h3><h3>7 Ns</h3><h3>Explanation</h3>

Given in the question,

mass of foul ball = 0.140 kg

initial speed with which ball was hit with the bat = 30 m/s

final speed = 40 m/s

According to the scenario the whole scene is making a right angle triangle

So, to the solve the question we will use pythagorus theorem

<h3>Hypotenuse² = base² + height²</h3>

Here,

Hypotenuse= Magnitude of impulse

Base = 1st change of momentum

height = 2nd change of momentum

1st impulse (1st change of momentum)

p = m(1)v(1) = (0.14 kg)(40.0 m/s) = 5.6 kg m / s = 5.6 N s

2nd impulse (2nd change of momentum)

p = m(2)v(2) = (0.14 kg)(30.0 m/s) = 4.2 kg m / s = 4.2 N s

Magnitude of impulse (hypotenuse of triangle)

impulse² = (5.6)² + (4.2)²

impulse² = 31.36 + 17.64

impulse² = 49

impulse² = √49

impulse = 7.0 N s

7 0

3 years ago

When light encounters a barrier with slits cut it in it, the light will bend through the slits creating a pattern like that seen

lawyer [7]

Answer:

reflection

Explanation:

an example would be looking in the mirror

8 0

3 years ago

Read 2 more answers

A block is released to slide down a frictionless incline of 15∘ and then it encounters a frictional surface with a coefficient o

Elodia [21]

The block's potential energy at the top of the incline (at a height h from the horizontal surface) is equal to its kinetic energy at the bottom of the incline, so that

mgh = 1/2 mv²

where v is its speed at the bottom of the incline. It follows that

v = √(2gh)

If the incline is 20.4 m long, that means the block has a starting height of

sin(15°) = h/(20.4 m) ⇒ h = (20.4 m) sin(15°) ≈ 5.2799 m

and so the block attains a speed of

v = √(2gh) ≈ 10.1728 m/s

The block then slides to a rest over a distance d. Kinetic friction exerts a magnitude F over this distance and performs an amount of work equal to Fd. By the work-energy theorem, this quantity is equal to the block's change in kinetic energy, so that

Fd = 0 - 1/2 mv² ⇒ d = (-1293.58 J)/F

By Newton's second law, the net vertical force on the block as it slides is

∑ F [vertical] = n - mg = 0

where n is the magnitude of the normal force, so that

n = mg = (25 kg) g = 245 N

and thus the magnitude of friction is

F = -0.16 (245 N) = -39.2 N

(negative since it opposes the block's motion)

Then the block slides a distance of

d = (-1293.58 J) / (-39.2 N) ≈ 32.9994 m ≈ 33 m

5 0

2 years ago