1998 - JAMB Physics Past Questions and Answers - page 1
1
A bullet fired at a wooden block of thickness 0.15m manages to penetrate the block. If the mass of the bullet is 0.025kg and the average resisting force of the wood is 7.5 x 103N, calculate the speed of the bullet just before it hits the wooden block.
A
450ms-1
B
400ms-1
C
300ms-1
D
250ms-1
correct option: c
F = ma
Therefore, a = \(\frac{F}{m}\)
a = \(\frac{7.5 \times 10^{3}}{0.025}\)
a = 3 x 105m/s2
using V2 = U2 + 2ax;
V2 = 02 + (2 x 3 x 105 x 0.15)
Therefore V = 300 m/s
Users' Answers & CommentsTherefore, a = \(\frac{F}{m}\)
a = \(\frac{7.5 \times 10^{3}}{0.025}\)
a = 3 x 105m/s2
using V2 = U2 + 2ax;
V2 = 02 + (2 x 3 x 105 x 0.15)
Therefore V = 300 m/s
2
The velocity of a sound wave at 27oC is 360ms-1. Its velocity at 127oC is
A
120√3 ms-1
B
240 ms-1
C
240√3 ms-1
D
720√3 ms-1
correct option: c
Velocity \(\alpha \sqrt{temperature}\)
\(\frac{V_1}{V_2} = \sqrt{\frac{T_1}{T_2}}\)
therefore, \(\frac{360}{V_2} = \sqrt{\frac{27 + 273}{127 + 273}}\)
\(V_2 = 240\sqrt{3}\)m/s
Users' Answers & Comments\(\frac{V_1}{V_2} = \sqrt{\frac{T_1}{T_2}}\)
therefore, \(\frac{360}{V_2} = \sqrt{\frac{27 + 273}{127 + 273}}\)
\(V_2 = 240\sqrt{3}\)m/s
3
The physical quantity that has the same dimensions as impulse is
A
energy
B
momentum
C
surface tension
D
pressure
4
A ball is moving at 18ms-1 in a direction inclined at 60o to the horizontal. The horizontal component to its velocity is
A
9√3ms-1 ·
B
6√3ms-1 ·
C
8√3ms-1 ·
D
9ms-1 ·
correct option: d
Users' Answers & Comments5
In free fall, a body of mass 1kg drops from a height of 125m from rest in 5s. How long will it take another body of mass 2kg to fall from rest from the same height? [g = 10ms-1·]
A
5s
B
10s
C
12s
D
15s
correct option: a
In free fall, every object will reach the ground at same time
Users' Answers & Comments6
A ball of mass 0.15kg is kicked against a rigid vertical wall with a horizontal velocity of 50ms-1. If it rebounced with a horizontal velocity of 30ms-1, Calculate the impulse of the ball on the ball · ·
A
3.0Ns
B
4.5Ns
C
7.5Ns
D
12.0Ns
7
The force of attraction between two point masses is 10-4N when the distance between them is 0.18m. If the distance is reduced to 0.06m, calculate the force.
A
1.1 x 10-5N
B
3.3 x 10-5N
C
3.0 x 10-4N
D
9.0 x 10-4N
correct option: d
F \(\alpha \frac{1}{r^2}\)
therefore, \(\frac{F_1}{F_2}\) = \(\frac{r_1}{(r_2)^2}\)
\(\frac{10^{-4}}{F_2}\) = \(\frac{(0.06)^2}{(0.18)^2}\)
F = 9 x 10-4N
Users' Answers & Commentstherefore, \(\frac{F_1}{F_2}\) = \(\frac{r_1}{(r_2)^2}\)
\(\frac{10^{-4}}{F_2}\) = \(\frac{(0.06)^2}{(0.18)^2}\)
F = 9 x 10-4N
8
A uniform meter rule weighing 0.5N is to be pivoted on a knife-edge at the 30cm-mark. where will a force of 2N be placed from the pivot to balance the meter rule?
A
95cm
B
25cm
C
20cm
D
5cm
correct option: b
Users' Answers & Comments9
A man whose mass is 80kg climbs a staircase in 20s and expends power of 120W. Find the height of the staircase.[g = 10ms-2]
A
1.8m
B
2.0m
C
2.5m
D
3.0m
correct option: d
Power x time = mgh
120 x 20 = 80 x 10 x h
2400 = 800h
making h the subject of formula,divide both side by 800
\(\frac{2400}{800}\) = h
h = 3m
Users' Answers & Comments120 x 20 = 80 x 10 x h
2400 = 800h
making h the subject of formula,divide both side by 800
\(\frac{2400}{800}\) = h
h = 3m
10
A parachute attains a terminal velocity when
A
its density is equal to the density of air
B
the viscous force of the air and the upthrust completely counteract its weight
C
it expands as a result of reduced external pressure
D
the viscous force of the air is equal to the sum of the weight and upthrust
correct option: b
Users' Answers & Comments