Ask question

Ask question

All categories

3 years ago

8

A block of mass 1.3 kg is resting at the base of a frictionless ramp. A bullet of mass 50 g is traveling parallel to the ramp su

rface at 250 m/s. It collides with the block, enters it, and exits the other side at 100 m/s. How far up the ramp will the block travel?

1 answer:

aleksandrvk [35]3 years ago

3 0

Answer:

s = 2.65 m

Explanation:

given,

mass of block,M = 1.3 Kg

mass of bullet,m = 50 g = 0.05 Kg

speed of bullet,u = 250 m/s

speed of bullet after collision,v = 100 m/s

distance traveled by the block = ?

Assuming the angle of inclination of ramp equal to 40°

calculating the speed of the block

using conservation of momentum

M u' + m u = m v + M v'

initial speed of the block is equal to zero

0 + 0.05 Kg x 250 = 0.05 x 100 + 1.3 x v'

1.3 v' = 7.5

v' = 5.77 m/s

now, calculation of acceleration

equation the horizontal component

-mg sin θ = ma

a = - g sin θ

a = - 9.8 x sin 40°

a = -6.29 m/s²

using equation of motion for the calculation of distance moved by the block

v² = u² + 2 a s

0² = 5.77² + 2 x (-6.29) x s

12.58 s = 33.29

s = 2.65 m

hence, the distance moved by the block is equal to 2.65 m

You might be interested in

Calculate the acceleration if you push with a 20-N horizontal force on a 2-kg block on a horizontal friction-free air table

Afina-wow [57]

Acceleration = force / mass = 20 / 2 = 10 m/s^2

3 0

3 years ago

A toy car having mass m = 1.10 kg collides inelastically with a toy train of mass M = 3.55 kg. Before the collision, the toy tra

kkurt [141]

Answer:

$V_{ft}= 317 cm/s$

ΔK = 2.45 J

Explanation:

a) Using the law of the conservation of the linear momentum:

$P_i = P_f$

Where:

$P_i=M_cV_{ic} + M_tV_{it}$

$P_f = M_cV_{fc} + M_tV_{ft}$

Now:

$M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}$

Where $M_c$ is the mass of the car, $V_{ic}$ is the initial velocity of the car, $M_t$ is the mass of train, $V_{fc}$ is the final velocity of the car and $V_{ft}$ is the final velocity of the train.

Replacing data:

$(1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}$

Solving for $V_{ft}$ :

$V_{ft}= 3.17 m/s$

Changed to cm/s, we get:

$V_{ft}= 3.17*100 = 317 cm/s$

b) The kinetic energy K is calculated as:

K = $\frac{1}{2}MV^2$

where M is the mass and V is the velocity.

So, the initial K is:

$K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2$

$K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2$

$K_i = 22.06 J$

And the final K is:

$K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2$

$K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2$

$K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2$

$K_f = 19.61 J$

Finally, the change in the total kinetic energy is:

ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J

4 0

3 years ago

A student touches sphere x and moves it close to, but not touching sphere y. What are the natures of the charges left on the two

e-lub [12.9K]

No charge I know this because

7 0

3 years ago

A certain runner averages 4.1 m/s over a 10

Ghella [55]

Answer: 40.650406504065 or 40 minutes and 39 seconds.

Explanation:

1 k = 1000m

race = 10000m

runner time = 10000 / 4.1

runner time = 2439.0243902439024 seconds

runner time = 2439.0243902439024/60 = 40.650406504065 or 40 minutes and 39 seconds.

6 0

3 years ago

Natalie lifts a 15-kg rock from the ground onto a 1.5 meter high wall. what is the amount of potential energy she has given the

Zina [86]

The amount of gravitational potential energy acquired by the rock is equal to:
$\Delta U = mg \Delta h$
where
m is the mass of the rock
g is the gravitational acceleration
$\Delta h$ is the increase in height of the rock

Substituting the data of the problem, we find
$\Delta U=(15 kg)(9.81 m/s^2)(1.5 m)=220.7 J$
So, Natalie gave 220.7 J of energy to the rock.

4 0

3 years ago