JAMB Physics Past Questions & Answers

131.

A car of mass 800kg attains a speed of 25m/s in 20secs. The power developed in the engine is?

1.25 x 10 $^{4}$ W

2.50x 10 $^{4}$ W

1.25 x 10 $^{6}$ W

2.50 x 10 $^{6}$ W

View answer & explanation

Correct answer is B

P = $\frac{mv^{2}}{t} = \frac{800 \times 25 \times 25}{20}$

2.5 x 10 $^{4}$ W

132.

A ball of mass 0.1kg is thrown vertically upwards with a speed of 10ms $^{-1}$ from the top of a tower 10m high. Neglecting air resistance, its total energy just before hitting the ground is? (Take g = 10ms $^{-2}$ )

10J

15J

20J

View answer & explanation

Correct answer is C

m = 0.1kg
u = 10ms $^{-1}$
h = 10m
g = 10ms $^{-2}$

At h $_{max}$ ; v = 0

Using
$\not v$ = $\not u$ = $\not gt$
$\not 0$ = $\not w$ - $\not 10t$

v $^{2}$ = u $^{2}$ - 2gh
0 $^{2}$ = 10 $^{2}$ - 2 x 10h
20h = 100
h = 5m
Total height = 10 + 5
= 15m
Total energy = mghr = 0.1 x 10 x1 5
= 15J

133.

A lead bullet of mass 0.05kg is fired with a velocity of 200ms $^{-1}$ into a block of mass 0.95kg. Given that the lead block can move freely, the final kinetic energy after impact is?

50J

100J

150J

200J

View answer & explanation

Correct answer is A

Given
m $_{1}$ = 0.05kg, u $_{1}$ = 200ms $^{-1}$ , m $_{2}$ = 0.95kg

K.E = $\frac{1}{2}$ m $_{r}$ v $^{2}$

m $_{1}$ u $_{1}$ = v(m $_{1}$ + m $_{2}$ ) [law of conversation of momentum]

v = $\frac{0.05 \times 200}{0.05 + 95}$ = 10ms $^{-1}$

K.E = $\frac{1}{2}$ (1)10 $^{2}$ = 50J

134.

Particles of mass 10 $^{-2}$ kg is fixed to the tip of a fan blade which rotates with angular velocity of 100rad $^{-1}$ . If the radius of the blade is 0.2m, the centripetal force is?

20N

200N

400N

View answer & explanation

Correct answer is B

Given
m = 10 $^{-2}$
w = 100rads $^{-1}$
F = mw $^{2}$ r
r = 0.2m
F = ?

= 10 $^{-2}$ x (100) $^{2}$ x 0.2

= 20N

135.

Two bodies have masses in the ratio3:1. They experience forces which impart to them, acceleration in the ratio 2:9 respectively. Find the ratio of forces the masses experienced

1:4

2:1

2:3

2:5

View answer & explanation

Correct answer is C

Given $\frac{m_{1}}{m_{2}} = \frac{3}{1}$ $\Rightarrow$ m $_{1}$ = 3m $_{2}$

$\frac{a_{1}}{a_{2}} = \frac{2}{9}$ = a $_{1}$ = $\frac{2a_{2}}{9}$

$\frac{F_{1}}{F_{2}}$ = ?

Since F = ma

$\Rightarrow$ $\frac{F_{1}}{F_{2}} = \frac{m_{1}a_{1}}{m_{2}a_{2}}$ = $\frac{(3 \not m_{2})(2 \not a_{2})}{\not m_{2} a_{2} \not a_{3}}$

$\frac{F_{1}}{F_{2}} = \frac{2}{3}$ $\Rightarrow$ 2:3

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