### From JAMB Physics past question 2017:

**Question:**

**Which of the following statements is correct?**A. The pressure exerted is higher with a narrow shoe than a flat heel shoe. B. No pressure is exerted in both narrow and flat heel shoes. C. The pressure exerted is lower with narrow heel shoes than a flat heel shoe. D. The pressure exerted is the same in both narrow and flat heel shoes.

**Solution:**Recall that the formula for pressure is: P = F/A, Where F = applied force and A = surface area. Hence, assuming the force is the same (and should be for this comparison), e have that: P = c/A, where c = constant (constant force). Hence, P is inversely proportional to A such that as A increases, P decreases and vice-versa. Since a narrow heel shoe has a smaller surface area (smaller A) than a flat heel shoe, it means that P is greater for a narrow heel shoe.

**Hence, at constant force (F), a narrow heel shoe will exert more pressure than a flat heel shoe. Answer: A**

### From JAMB Physics past question 2013:

**Question: When a brick is taken from the earth’s**

**surface to the moon, its mass**A. Becomes zero B. Remains constant C. Reduces D. increases

**Explanation:**Unlike weight which varies with the acceleration due to gravity for a given mass, the mass of an object does not change with location or place. Recall that in the formula for weight,

**w = mg, where m = constant.**The value of

**g**will change from one location to another within the earth and,

**in the moon, the value of g drops to about 16m/s**. These changes in the value of g lead to changes in the weight of the object. However, the changes, which affects the weight, do not affect the mass of the given object.

^{2}**Answer: B**

### From JAMB Physics past question 2013:

**Question: A simple pendulum of length 0.4m has a period 2s. What is the period of a similar pendulum of length 0.8m at the same place?**A. √2s B. 8s C. 4s D. 2√2s

**Solution:**

**Given:**l

_{1}= 0.4m T

_{1}= 2s l

_{2}= 0.8m

**Find**: T

_{2}The pendulum in the question performs Simple Harmonic Motion (SHM). The formula for finding the period (

**T**)

**of such motion is:**

**T = 2π√(l/g)**Where: T = period of oscillation, l = length of the pendulum. g = acceleration due to gravity = 10m/s

^{2 }(assumed approximate value). The next step is to factor out the constants (2, π and g) from the equation: » Square both sides: T

^{2}= (2π)

^{2}(l/g) » T

^{2}= ((2π)

^{2})/g) * l Since 2, π and g are constants, we can bundle them together and call them one constant, say k. Hence, » T

^{2}= kl, where k = (2π)

^{2})/g T

^{2}= kl T

^{2}/l = k Let's use subscript 1 to denote the first pendulum and subscript 2 to denote the second, hence, T

_{1}

^{2}/l

_{1}= k, also T

_{2}

^{2}/l

_{2}= k Hence, T

_{1}

^{2}/l

_{1}= T

_{2}

^{2}/l

_{2}= k » T

_{2}

^{2}/T

_{1}

^{2}= l

_{2}/l

_{1}» T

_{2}

^{2}/(2)

^{2}= (0.8)/(0.4) » T

_{2}

^{2}/4 = 2 » T

_{2}

^{2}/4 = 2*4 = 8 » T

_{2}= √8 = √(4

*2) = √(4)*√(2) = 2√(2) Hence, T

_{2}= 2√2s

**Answer: D**

### From JAMB Physics past question 2013:

A train with an initial velocity of 20ms^{-1}is subjected to a uniform deceleration of 2ms

^{-2}. The time required to bring the train to a complete halt is A. 40s B. 5s C. 10s D. 20s

**Solution**

**Given:**u = 20ms

^{-1}-a = 2ms

^{-1}(Note: -a because it is deceleration)

**Find: v**From Newton's law of linear motions,

**v = u + at**Where t = time, u = initial velocity, and v = final velocity. By the time the train is brought to a complete halt,

**v = 0**Therefore, 0 = u + at at = -u t = -(u/a) But -a = 2 implies that a = -2 Hence, t = -(20/-2) = 20/2 = 10 t = 10s.

**Answer: C**

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