All categories

1 year ago

A 0.0427 kg racquet-ball is moving

Physics

1 answer:

Gwar [14]1 year ago

4 0

Answer:

Mass of the box = 0.9433 kg

Explanation:

Mass of racket-ball $(m_1)$ = 0.00427 kg

Velocity of racket-ball before collision $(v_{1i})$ = 22.3 m/s

Velocity of racket-ball after collision with box $(v_{1f})$ = -11.5 m/s

[Since ball is bouncing back, so velocity is taken negative.]

Velocity of the box before collision $v_{2i}$ = 0 m/s

<em>[Since the box is stationary, so velocity is taken zero]</em>

Velocity of box moving forward after collision $v_{2f}$ = 1.53 m/s

To find the mas of the box $m_2$ .

By law of conservation of momentum we have:

Momentum before collision = Momentum after collision

This can be written as:

$p_i=p_f$

$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$

We can plugin the given value to find $m_2$

$(0.0427\times 22.3)+(m_2\times 0)=(0.0427\times (-11.5))(m_2\times 1.53)$

$0.9522+0=-0.4911+1.53m_2$

Adding both sides by 0.4911

$0.9522+0.4911=-0.4911+0.4911+1.53m_2$

$1.4433=1.53m_2$

Dividing both sides by 1.53.

$\frac{1.4433}{1.53}=\frac{1.53m_2}{1.53}$

$0.9433=m_2$

∴ $m_2=0.9433$ kg

Mass of the box = 0.9433 kg (Answer)

You might be interested in

A 20kg mass approaches a spring at a speed of 30 m/s. The mass compresses the spring 12cm before coming to a stop. Calculate the

Oksana_A [137]

Answer:

625000 N/ m

Explanation:

m= 20 kg

v= 30 m/s

x= 12 cm

k = ?

Here when the mass when hits at spring its speed is

Vi= 30 m/s

Finally it comes to rest after compressing for 12 cm

i-e Vf = 0 m/s

Distance= S= 12 cm = 0.12 m

using

2aS= Vf2 - Vi2

==> 2a ×0.12 = o- 30 × 30

==> a = 900 ÷ 0.24 = 3750 m/sec2

Now we know;

F = ma

F= -Kx

==> ma= -kx

==> 20 × 3750 = -K × 0.12

==> k = 625000 N/ m

5 0

1 year ago

A large crate is suspended from the end of a vertical rope. Is the tension in the rope greater when the crate is at rest or when

choli [55]

Answer:

Part a)

the tension force is equal to the weight of the crate

Part b)

tension force is more than the weight of the crate while accelerating upwards

tension force is less than the weight of crate if it is accelerating downwards

Explanation:

Part a)

When large crate is suspended at rest or moving with uniform speed then it is given as

$F_t - mg = ma$

here since speed is constant or it is at rest

so we will have

$a = 0$

$F_t = mg$

so the tension force is equal to the weight of the crate

Part b)

Now let say the crate is accelerating upwards

now we can say

$F_t - mg = ma$

$F_t = mg + ma$

so tension force is more than the weight of the crate

Now if the crate is accelerating downwards

$F_t - mg = -ma$

$F_t = mg - ma$

so tension force is less than the weight of crate if it is accelerating downwards

4 0

2 years ago

You throw a baseball at an angle of 30.0∘∘ above the horizontal. It reaches the highest point of its trajectory 1.05 ss later. A

garri49 [273]

Answer:

The speed with which the baseball leaves the hand = 20.58 m/s

Explanation:

The time take to reach highest height during a projectile's flight is given by

t = (u sin θ)/g

u = initial velocity of the baseball = ?

θ = angle of throw above the horizontal

g = acceleration due to gravity = 9.8 m/s²

1.05 = (u sin 30)/9.8

u = (1.05 × 9.8)/0.5

u = 20.58 m/s

7 0

1 year ago

Calculate the amount of work done to draw a current of 8A from a point at 100V to a point at 120V in 2 seconds?

Morgarella [4.7K]

Given:
I=8A
t=2second
Potential difference,V=120-100=20volt
Workdone=V×i×t
=20×8×2
=320 joule.

3 0

1 year ago

Consider a spring that does not obey Hooke’s law very faithfully. One end of the spring is fixed. To keep the spring stretched o

IRINA_888 [86]

Answer:

a) $W=-0.0103125\ J$

b) $W=0.0059375\ J$

c) Compressing is easier

Explanation:

Given:

Expression of force:

$F=kx-bx^2+cx^3$

where:

$k=100\ N.m^{-1}$

$b=700\ N.m^{-2}$

$c=12000\ N.m^{-3}$

$x$ when the spring is stretched

$x$ when the spring is compressed

hence,

$F=100x-700x^2+12000x^3$

From the work energy equivalence the work done is equal to the spring potential energy:

here the spring is stretched so, $x=-0.05\ m$

Now,

The spring constant at this instant:

$j=\frac{F}{x}$

$j=\frac{100\times (-0.05)-700\times (-0.05)^2+12000\times (-0.05)^3}{-0.05}$

$j=-8.25\ N.m^{-1}$

Now work done:

$W=\frac{1}{2} j.x^2$

$W=0.5\times -8.25\times (-0.05)^2$

$W=-0.0103125\ J$

When compressing the spring by 0.05 m

we have, $x=0.05\ m$

<u>The spring constant at this instant:</u>

$j=\frac{F}{x}$

$j=\frac{100\times (0.05)-700\times (0.05)^2+12000\times (0.05)^3}{0.05}$

$j=4.75\ N.m^{-1}$

Now work done:

$W=\frac{1}{2} j.x^2$

$W=0.5\times 4.75\times (0.05)^2$

$W=0.0059375\ J$

Since the work done in case of stretching the spring is greater in magnitude than the work done in compressing the spring through the same deflection. So, the compression of the spring is easier than its stretching.

8 0

1 year ago