1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
goblinko [34]
3 years ago
12

Blocks A (mass 3.50 kg) and B (mass 6.50 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block

A is moving toward it at 2.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line. Let +x be the direction of the initial motion of block A.a) Find the velocity of block A when the energy stored in the spring bumpers is maximum.b) Find the velocity of block A when the energy stored in the spring bumpers is maximum.c) Find the velocity of block B when the energy stored in the spring bumpers is maximum.d) Find the velocity of block A after they have moved apart.e) Find the velocity of block B after they have moved apart.
Physics
1 answer:
Vedmedyk [2.9K]3 years ago
5 0

Answer:

(a) V (A) =  0.7 m/s,

(b) V (A) =  0.7 m/s,

(c) V (B) =  0.7 m/s

(d) u= - 0.60 m/s

(e) v = 0.75 m/s

Explanation:

Given:

M(A) =3.50 Kg, M(B)=6.50 Kg, V(A) = 2.00 m/s, V(B) = 0 m/s

Sol:

a)  law of conservation of momentum

M(a) x V(A) + M(B) x V(B) = ( M(a) + M(B) ) V      (let V is Common Velocity of Both block)

so 3.50 Kg x 2.00 m/s + 6.50 Kg x 0 m/s = (3.50 Kg + 6.50 Kg ) V

after solving V =  0.7 m/s

After the collision the velocities of the both block will be as the the spring is compressed maximum.

V (A) =  0.7 m/s

b)  V(A) =  0.7 m/s ( Part (a) and Part (a) are repeated )

c) as stated above the in the Part (a)

V(B) =  0.7 m/s

d) When the both blocks moved apart after the collision:

Let u=velocity of block A after the collision.

and v = velocity of block B after the collision.

then conservation of momentum

M(a) x V(A) + M(B) x V(B) = M(a) x v + M(B) x u

⇒ 3.50 Kg x 2.00 m/s + 6.50 Kg x 0 m/s =  3.50 Kg x u + 6.50 Kg x v

⇒ 2.00 m/s = u + 1.86 v -----eqn (1)  ( dividing both side by 3.50 Kg)

For elastic collision  

the velocity relative approach = velocity relative separation

so 2.00 m/s = v-u  ----- eqn (2)

⇒v = u + 2.00 m/s

putting this value in eqn (1) we get

2.00 m/s = u + 1.86 (v + 2.00 m/s)

u= - 0.60 m/s

e) putting v= 2.00 m/s in eqn (1)

2.00 m/s = - 2.32 m/s + 1.86 v

v = 0.75 m/s

You might be interested in
An automobile engine delivers 47.4 hp. how much time will it take for the engine to do 6.82 × 105 j of work? one horsepower is e
kogti [31]
1 horsepower is equal to 746 W, so the power of the engine is
P=47.4 hp \cdot 746  \frac{W}{hp}=35360 W
The power is also defined as the energy E per unit of time t:
P= \frac{E}{t}
Where the energy corresponds to the work done by the engine, which is E=6.82 \cdot 10^5 J. Re-arranging the formula, we can calculate the time t needed to do this amount of work:
t =  \frac{E}{P}= \frac{6.82 \cdot 10^5 J}{35360 W}=19.3 s
8 0
3 years ago
You are given a parallel plate capacitor that has plates of area 29 cm2 which are separated by 0.0100 mm of nylon (dielectric co
GalinKa [24]

Answer:

20.60 kV

Explanation:

Capacitance of parallel plates without dielectric between them is:

C=\frac{\varepsilon_{0}A}{d}

with d the distance between the plates, A the area of the plates and ε₀ the constant 8.85419\times10^{-12}\frac{C^{2}}{Nm^{2}}, so :

C_0=\frac{(8.85419\times10^{-12})(0.0029)}{0.0100\times10^{-3}}=2.57\times10^{-9} F

But the dielectric constant is defined as:

k=\frac{C}{C_{0}}

With C the effective capacitance (with the dielectric) and Co the original capacitance (without the dielectric). So, the new capacitance is:

C=kC_0

But capacitance is related with voltage by:

C=\frac{Q}{V}

with Q the charge and V the voltage, using the new capacitance and solving for V:

kC_0=\frac{Q}{V}

V=\frac{Q}{kC_0}=\frac{0.18\times10{-3}}{(3.4)(2.57\times10^{-9})}=20599.68V=20.60 kV

4 0
3 years ago
Read 2 more answers
B. Why does the moon revolve around the earth? Give<br>reason.[2]​
LenKa [72]

Answer:

The moon revolves around Earth because Earth is larger than the moon, so it is heavier, and has a greater gravitational pull. The plane of the moon's orbit is very close to the plane of Earth's orbit around the Sun. This is why planets revolve around the Sun, because it is larger, so therefore it has a greater gravitational pull.

6 0
3 years ago
A bullet with a mass ????=12.5 g and speed ????=86.4 m/s is fired into a wooden block with ????=113 g which is initially at rest
skad [1K]

Answer:

a) 8.61 m/s, b) 5.73 m

Explanation:

a) During the collision, momentum is conserved.

mv = (m + M) V

(12.5 g) (86.4 m/s) = (12.5 g + 113 g) V

V = 8.61 m/s

b) After the collision, energy is conserved.

Kinetic energy = Work done by friction

1/2 (m + M) V² = F d

1/2 (m + M) V² = N μk d

1/2 (m + M) V² = (m + M) g μk d

1/2 V² = g μk d

d = V² / (2g μk)

d = (8.61 m/s)² / (2 × 9.8 m/s² × 0.659)

d = 5.73 m

Notice we used the kinetic coefficient of friction.  That's the friction when an object is moving.  The static coefficient of friction is the friction on a stationary object.  Since the bullet/block combination is sliding across the surface, we use the kinetic coefficient.

4 0
3 years ago
A boat heads north in still water at 4.5 m/s directly across a river that is running east at 3.0 m/s. What is the velocity of th
Fynjy0 [20]
<h2>Velocity of the boat with respect to Earth is 5 m/s 56.31° north of east</h2>

Explanation:

A boat heads north in still water at 4.5 m/s directly across a river that is running east at 3.0 m/s.

Let north be positive y axis and east be positive x axis.

We have

         Velocity of boat = 4.5 j m/s

          Velocity of river = 3 i m/s

Velocity of the boat with respect to Earth = 3 i + 4.5 j

\texttt{Magnitude of velocity = }\sqrt{3^2+4.5^2}\\\\\texttt{Magnitude of velocity = }5.41m/s

\texttt{Angle = }tan^{-1}\left ( \frac{4.5}{3}\right )\\\\\texttt{Angle = }56.31^0

That is 56.31° north of east.

Velocity of the boat with respect to Earth is 5 m/s 56.31° north of east

4 0
3 years ago
Read 2 more answers
Other questions:
  • If the latent heat of fusion of ice is 334kj/kg and the specific heat of water is 4.18kj kg-1 k-1 how much heat, in joules, must
    9·1 answer
  • If blue light hits a red filter, what kind of light comes through the filter?
    7·1 answer
  • An object is pulled with two forces, 10 N northward and 15 N southward. The direction of the net force is to the An object is pu
    9·1 answer
  • A Carnot engine takes 3000 J from a reservoir at 600 K, does some work, and discards some heat to a reservoir at 300 K. What is
    9·1 answer
  • Consider two diffraction gratings. One grating has 3000 lines per cm, and the other one has 6000 lines per cm. Both gratings are
    5·1 answer
  • Which of the following is a result of the transfer of energy?
    9·2 answers
  • What is nuclear energy
    5·1 answer
  • 1.) What is the equation for Average Speed?
    10·2 answers
  • Shanika is an engineer at an amusement park who is experimenting with changes to the setup for a magnetic roller coaster ride. I
    6·1 answer
  • Forces present on a flying dove
    10·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!