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Test Prep MCAT Test Exam
Medical College Admission Test: Verbal Reasoning, Biological Sciences, Physical Sciences, Writing Sample (Page 20 )

Updated On: 30-Jan-2026
Page 20 of 164
PDF with 811 Questions

Which of the following is true when ice melts?

  1. The changes in both enthalpy and entropy are positive.
  2. The changes in both enthalpy and entropy are negative.
  3. The charge in enthalpy is positive; the change in entropy is negative.
  4. The change in enthalpy is negative; the change in entropy is positive.

Answer(s): A

Explanation:

When heat is absorbed during a reaction, H is positive; when heat is released, H is negative. The melting of ice requires absorption of heat to disrupt the attractions between molecules, so the change in enthalpy is
positive, and you can eliminate choice B and choice D. Now, entropy is the measurement of the disorder in a system; the change in entropy is expressed as S. When the disorder of a system increases, S is positive; when disorder decreases, S is negative. When ice melts, the individual molecules of water become freer to move around, so there is an increase in disorder and S is positive. Therefore, choice C is wrong and choice A is the correct answer.



When softball players take batting practice, they often use a machine called an "automatic pitcher," which is essentially a cannon that uses air pressure to launch a projectile. In a prototype automatic pitcher, a softball is loaded into the barrel of the cannon and rests against a flat disk. That disk is locked into place, and a high air pressure is built up behind it. When the disk is released, the softball is pushed along the barrel of the cannon and ejected at a speed of V0.
Figure 1 shows the batter and automatic pitcher. The angle of the barrel to the horizontal is . The unit vectors i and j point in the horizontal and vertical directions respectively.

Figure 1
The height above the ground y of the softball as a function of time t is shown in Figure 2, where t = 0 at Point A, t = tB at Point B, and t = tC at Point C. The softball is ejected from the barrel of the cannon at Point A; it reaches its maximum height at Point B; and the batter hits the softball at Point C. (Note: Assume that the effects of air resistance are negligible unless otherwise stated.)

Figure 2
What physical quantity is NOT the same at Point C as at Point A?

  1. The velocity of the softball
  2. The speed of the softball
  3. The gravitational potential energy of the softball
  4. The horizontal component of the velocity of the softball

Answer(s): A

Explanation:

The velocity of the softball is a vector, which means that it has both a magnitude and direction. At Point A, the velocity vector points up and to the right, because the softball is traveling in his direction. At Point C, however, the velocity vector points down and to the right. Since the velocity's direction has changed, it is not the same at Points A and C, and choice A is correct.
You should know that the gravitational potential energy of the softball is given by U = mgy, where m is the mass, g is the acceleration due to gravity, and y is the height above the Earth. Since Points A and C are at the same height y, the gravitational potential is the same and choice C is incorrect.
The passage says to ignore air resistance, so the total mechanical energy, which is the kinetic plus potential energy, is conserved in this system. Since the gravitational potential is the same at Points A and C, the kinetic energy (1/2)mv2 must be the same as well. Because the mass m does not change, the speed v must be the same at both points also, and choice B is wrong.
Since air resistance is negligible, the only force present is that of gravity, which acts in the vertical direction.
Consequently, there is no horizontal force on the softball, which means there can be no horizontal acceleration.
Therefore, the horizontal component of the velocity must remain the same throughout the softball's flight, and choice D is wrong.



When softball players take batting practice, they often use a machine called an "automatic pitcher," which is essentially a cannon that uses air pressure to launch a projectile. In a prototype automatic pitcher, a softball is loaded into the barrel of the cannon and rests against a flat disk. That disk is locked into place, and a high air pressure is built up behind it. When the disk is released, the softball is pushed along the barrel of the cannon and ejected at a speed of V0.
Figure 1 shows the batter and automatic pitcher. The angle of the barrel to the horizontal is . The unit vectors i and j point in the horizontal and vertical directions respectively.

Figure 1
The height above the ground y of the softball as a function of time t is shown in Figure 2, where t = 0 at Point A, t = tB at Point B, and t = tC at Point C. The softball is ejected from the barrel of the cannon at Point A; it reaches its maximum height at Point B; and the batter hits the softball at Point C. (Note: Assume that the effects of air resistance are negligible unless otherwise stated.)


Figure 2
What is the acceleration of the softball t seconds after it exits the barrel?

  1. -gj
  2. -v0/ti
  3. -v0/tj
  4. -v0 / (ti ­ gj)

Answer(s): A

Explanation:

Newton's second law F = ma states that a force produces an acceleration in the same direction as the force. In the absence of a force, there is no acceleration. In this situation, the only force present is the force of gravity, which is the same at all times during the softball's flight. Near the Earth's surface, the gravitational force is directed downward and its magnitude is given by F = mg, where m is the softball's mass and g is the acceleration due to gravity. Putting the gravitational force into Newton's second law, we obtain ma = mg.
Canceling the masses, we obtain a = g. In other words, the acceleration of the softball is the same as the acceleration due to gravity.
Figure 1 shows that the unit vector j points upwards, which means that -j points downwards.
Because j is a unit vector, its magnitude is 1. Since the softball accelerates downward with a magnitude of g, the acceleration is symbolically given by g(-j) = -gj. Thus, choice A is correct. Choice C is incorrect because it has the wrong magnitude. Choices B and D are wrong because they both include an acceleration in the horizontal direction -i.



When softball players take batting practice, they often use a machine called an "automatic pitcher," which is essentially a cannon that uses air pressure to launch a projectile. In a prototype automatic pitcher, a softball is loaded into the barrel of the cannon and rests against a flat disk. That disk is locked into place, and a high air pressure is built up behind it. When the disk is released, the softball is pushed along the barrel of the cannon and ejected at a speed of V0.
Figure 1 shows the batter and automatic pitcher. The angle of the barrel to the horizontal is . The unit vectors i and j point in the horizontal and vertical directions respectively.


Figure 1
The height above the ground y of the softball as a function of time t is shown in Figure 2, where t = 0 at Point A, t = tB at Point B, and t = tC at Point C. The softball is ejected from the barrel of the cannon at Point A; it reaches its maximum height at Point B; and the batter hits the softball at Point C. (Note: Assume that the effects of air resistance are negligible unless otherwise stated.)

Figure 2
How will V0 change if the impulse on the softball remains the same but its mass is doubled?

  1. It will decrease by a factor of 4.
  2. It will decrease by a factor of 2.
  3. It will not change.
  4. It will increase by a factor of 2.

Answer(s): B

Explanation:

You should know that the change in momentum of a body is equal to the impulse applied to that body. The magnitude of the momentum of a body is given by mv, where m is its mass and v is its speed. In this situation an impulse is applied to the softball to launch it from the cannon. Before the impulse the speed of the softball is zero, so we can write J = (mv) = mv0 - m(0) = mv0, where J is the impulse. Solving for V0 , we obtain V0 = J/m.
Since m is inversely proportional to V0 , doubling the mass will halve V0 , so choice B is correct.



When softball players take batting practice, they often use a machine called an "automatic pitcher," which is essentially a cannon that uses air pressure to launch a projectile. In a prototype automatic pitcher, a softball is loaded into the barrel of the cannon and rests against a flat disk. That disk is locked into place, and a high air pressure is built up behind it. When the disk is released, the softball is pushed along the barrel of the cannon and ejected at a speed of V0.
Figure 1 shows the batter and automatic pitcher. The angle of the barrel to the horizontal is . The unit vectors i and j point in the horizontal and vertical directions respectively.

Figure 1
The height above the ground y of the softball as a function of time t is shown in Figure 2, where t = 0 at Point A, t = tB at Point B, and t = tC at Point C. The softball is ejected from the barrel of the cannon at Point A; it reaches its maximum height at Point B; and the batter hits the softball at Point C. (Note: Assume that the effects of air resistance are negligible unless otherwise stated.)

Figure 2
What is the ratio of the horizontal distance traveled by the softball at Point B to the horizontal distance traveled at Point C?

  1. 5:1
  2. 4:1
  3. 3:3
  4. 1:2

Answer(s): D

Explanation:

Figure 2 shows that the curve is symmetric around t = tB. Since the axes of graphs are always linear unless otherwise stated, the time elapsed at Point C is twice the time elapsed at Point B.
There are no horizontal forces because air resistance is negligible. Therefore, the horizontal speed is the constant V0 throughout the softball's flight. The horizontal distance traveled is then given by x = V0t, where t is the time elapsed. Since the time elapsed at Point C is twice the time elapsed at Point B, the horizontal distance traveled at Point C is twice the horizontal distance traveled at Point B, and choice D is correct.



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