All categories

2 years ago

A 2 kg ball of clay moving at 35 m/s strikes a 10 kg box initially at rest. What is the velocity of the box after the collision?

Physics

1 answer:

Ainat [17]2 years ago

4 0

Answer:

V = 5.83 m/s

Explanation:

Given that,

Mass of a ball of a clay, m = 2 kg

Initial speed of the clay, u = 35 m/s

Mass of a box, m' = 10 kg

Initially, the box was at rest, u' = 0

We need to find the velocity of the box after the collision. Let V be the common speed. Using the conservation of momentum to find it.

$mu+m'u'=(m+m')V\\\\V=\dfrac{mu+m'u'}{(m+m')}\\\\V=\dfrac{2\times 35+10\times 0}{2+10}\\\\V=5.83\ m/s$

So, the velocity of the box after the collision is equal to 5.83 m/s.

You might be interested in

Sam, whose mass is 78 kg , stands at the top of a 11-m-high, 110-m-long snow-covered slope. His skis have a coefficient of kinet

Valentin [98]

Answer:

v = 8.09 m/s

Explanation:

For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.

Let's calculate the energy

starting point. Higher

Em₀ = U = m gh

final point. To go down the slope

Em_f = K = ½ m v²

The work of the friction force is

W = fr L cos 180

to find the friction force let's use Newton's second law

Axis y

N - W_y = 0

N = W_y

X axis

Wₓ - fr = ma

let's use trigonometry

sin θ = y / L

sin θ = 11/110 = 0.1

θ = sin⁻¹ 0.1

θ = 5.74º

sin 5.74 = Wₓ / W

cos 5.74 = W_y / W

Wₓ = W sin 5.74

W_y = W cos 5.74

the formula for the friction force is

fr = μ N

fr = μ W cos θ

Work is friction force is

W_fr = - μ W L cos θ

Let's use the relationship of work with energy

W + ΔU = ΔK

-μ mg L cos 5.74 + (mgh - 0) = 0 - ½ m v²

v² = - 2 μ g L cos 5.74 +2 (gh)

v² = 2gh - 2 μ gL cos 5.74

let's calculate

v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74

v² = 215.6 -150.16

v = √65.44

v = 8.09 m/s

6 0

2 years ago

Calculate the force of gravity on the 0.60-kg mass if it were 1.3×107 m above Earth's surface (that is, if it were three Earth r

KIM [24]

The force of gravity between two objects is given by:
$F=G \frac{m_1 m_2}{r^2}$
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation

In this problem, the mass of the object is $m_1=0.60 kg$ , while the Earth's mass is $m_2=5.97 \cdot 10^{24} kg$ . Their separation is $r=1.3 \cdot 10^7 m$ , therefore the gravitational force exerted on the object is
$F=(6.67 \cdot 10^{-11}m^3 kg^{-1} s^{-2}) \frac{(0.60 kg)(5.97 \cdot 10^{24} kg)}{(1.3 \cdot 10^7 m)^2}=1.4 N$

5 0

2 years ago

A car has a kinetic energy of 1.9 × 10^3 joules. If the velocity of the car is decreased by half, what is its kinetic energy?

VLD [36.1K]

The initial kinetic energy of the car is
$E_1 = \frac{1}{2}mv_1^2 = 1.9 \cdot 10^3 J$

Then, the velocity of the car is decreased by half: $v_2 = \frac{v_1}{2}$
so, the new kinetic energy is
$E_2 = \frac{1}{2}mv_2 ^2 = \frac{1}{2} m ( \frac{v_1}{2} )^2= \frac{1}{2}m \frac{v_1^2}{4}= \frac{E_1}{4}$
So, the new kinetic energy is 1/4 of the initial kinetic energy of the car. Numerically:
$E_2 = \frac{1.9 \cdot 10^3 J}{4}=475 J$

5 0

3 years ago

1pt A cannon fires a 5-kg ball horizontally from a

Klio2033 [76]

Answer: Both cannonballs will hit the ground at the same time.

Explanation:

Suppose that a given object is on the air. The only force acting on the object (if we ignore air friction and such) will be the gravitational force.

then the acceleration equation is only on the vertical axis, and can be written as:

a(t) = -(9.8 m/s^2)

Now, to get the vertical velocity equation, we need to integrate over time.

v(t) = -(9.8 m/s^2)*t + v0

Where v0 is the initial velocity of the object in the vertical axis.

if the object is dropped (or it only has initial velocity on the horizontal axis) then v0 = 0m/s

and:

v(t) = -(9.8 m/s^2)*t

Now, if two objects are initially at the same height (both cannonballs start 1 m above the ground)

And both objects have the same vertical velocity, we can conclude that both objects will hit the ground at the same time.

You can notice that the fact that one ball is fired horizontally and the other is only dropped does not affect this, because we only analyze the vertical problem, not the horizontal one. (This is something useful to remember, we can separate the vertical and horizontal movement in these type of problems)

7 0

3 years ago

What is a learned behavior of a puppy

Ludmilka [50]

Its when the animal repeats what it is taught similar to a child you never really teach them to lie it just happens because they have seen other people do it

3 0

3 years ago