A boy throws a 15 kg ball at 4.7 m/s to a 65 kg girl who is stationary and standing on a skateboard. After catching the ball, the girl is travelling at: a) 0.88 m/s b) 0 m/s c) 1.1 m/s d) 3.2 m/s


Answer 1




To solve this problem we use the Momentum's conservation Law, before and after the girl catch the ball:

\n p_(1)=p_(2)\nm_(ball)*v_(o.ball)+m_(girl)*v_(o.girl) = m_(ball)*v_(f.ball) + m_(girl)*v_(f.girl)        (1)

At the beginning the girl is  stationary:

v_(o.girl)=0m/s       (2)

If the girl catch the ball, both have the same speed:

v_(f.girl)=v_(f.ball)=v_(f)       (3)

We replace (2) and (3) in (1):

m_(ball)*v_(o.ball) = (m_(ball)+m_(girl))*v_(f) \n

We can now solve the equation for v_{f}:


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At a given instant the bottom A of the ladder has an acceleration aA = 4 f t/s2 and velocity vA = 6 f t/s, both acting to the left. Determine the acceleration of the top of the ladder, B, and the ladder’s angular acceleration at this same instant.

The surface pressure of the atmosphere is about 14.7 psi (pounds per square inch). How many pounds per square yard does that amount to



14.7 psi is equal to 19051.2 pounds per square yard.


Dimensionally speaking, a square yard equals 1296 square inches. Therefore, we need to multiply the atmospheric pressure by 1296 to obtain its equivalent in pounds per square yard. That is:

p = 14.7\,(lbf)/(in^(2))* 1296\,(in^(2))/(yd^(2))

p = 19051.2\,(lbf)/(yd^(2))

14.7 psi is equal to 19051.2 pounds per square yard.

Two blocks of masses 1 = 700 and 2 = 1100 are connected by a cord of negligible mass and hung over a diskshaped pulley, as shown in the figure. The pulley has a mass of = 1.50 and a radius of = 14 , and rotates about a lightweight axle through its center. The axle itself is hung from the ceiling by two like cords of negligible mass and is held horizontally. The system is released from rest. a) Draw a free-body diagram for each of the blocks and the pulley separately. b) Find the magnitude of the acceleration of the blocks. c) Find the magnitude of the angular acceleration of the pulley. d) Find the magnitude of tensions in the cords, 1, 2, and 3. (See the figure.)



b) 16 cm

Magnification, m = v/u

3 = v/u

⇒ v = 3u

Lens formula : 1/v – 1/u = 1/f

1/3u = 1/u = 1/12

-2/3u = 1/12

⇒ u = -8 cm

V = 3 × (-8) = -24

Distance between object and image = u – v = -8 – (-24) = -8 + 24 = 16 cm


An internal explosion breaks an object, initially at rest,intotwo pieces, one of which has 1.5 times the mass of the other.If
7500 J were released in the explosion, how much kinetic energydid
each piece acquire?



4500 J and 3000 J


According to conservation of momentum

      0 = m_1 V_1 + m_2 V_2

Given that m_2 = 1.5 m_1 , so

    V_1 = -1.5 V_2

  the kinetic energy of each piece is

    K_2= (1)/(2) m_2v_2^2

    K_1= (1)/(2) m_1v_1^2

substituting the value of V1 in the above equation

    K_1 = (1/2)( m_2 / 1.5 )( -1.5 V_2)^2 = 1.5 (1/2)m_2 V_2^2 = 1.5 K_2

  Given that

         K_1 + k_2 = 7500 J

       1.5 K_2 + K_2 = 7500

         K_2 = 7500 / 2.5

               = 3000 J

this is the KE of heavier mass

      K_1 = 7500 - 3000 = 4500 J

this is the KE of lighter mass

Final answer:

The question is about finding the kinetic energy acquired by each of two pieces of an object following an internal explosion, using principles of conservation of energy and momentum in physics.


The student has asked about an internal explosion that breaks an object into two pieces with different masses, releasing a certain amount of kinetic energy in the process. This question involves applying the principle of conservation of energy and momentum to find the kinetic energy acquired by each piece post-explosion.

Assuming piece 1 has a mass of m and piece 2 has a mass of 1.5m, the total mass of the system is 2.5m. Since 7500 J of energy was released in the explosion, to find the kinetic energy of each piece, we can use the fact that the total kinetic energy is equal to the energy released during the explosion. Let the kinetic energy of the smaller piece be K1 and of the larger piece be K2. Because the object was initially at rest and momentum must be conserved, the momenta of the two pieces must be equal and opposite. This relationship allows us to derive the ratio of the kinetic energies. We can solve for K1 and K2 proportionally. Finally, because the kinetic energy is a scalar quantity, adding the kinetic energies of the two pieces will equal the total energy released.

Learn more about Kinetic Energy Acquisition here:



How to Measure the mass of a coin using a triple beam balance




A cylinder with a diameter of 2.0 in. and height of 3 in. solidifies in 3 minutes in a sand casting operation. What is the solidification time if the cylinder height is doubled? What is the time if the diameter is doubled?



3 min 55 sec is the solidification time if the cylinder height is doubled

7min 40 sec if the diameter is doubled


see the attachment

Which of the following could be reasonable explanations for how a piece of invisible tape gets charged? Select all that apply. (1) Protons are pulled out of nuclei in one tape and transferred to another tape.
(2) Charged molecular fragments are broken off one tape and transferred to another.
(3) Electrons are pulled out of molecules in one tape and transferred to another tape.
(4) Neutrons are pulled out of nuclei in one tape and transferred to another tape.



2 and 3


The right answer is the option 2 and 3,

This is all about the electrons transfer from one material to the other material.

For example if the electrons in the valence shell of one material are loosely attached then the other material's atoms try to take those electron to complete their shells and that is how the charges transfers from one another.

And it could also be happen as in option 2.