JAMB Physics Past Questions & Answers

1,781.

A body of mass 12kg traveling at 4.2ms-1 collides with a second body of mass 18kg at rest. Calculate their common velocity if the two bodies coalesce after collision?

1.5 ms^-1

1.4 ms^-1

2.1 ms^-1

1.7 ms^-1

View answer & explanation

Correct answer is D

Sum of momentum before collision
M1 U1 + M2M2
12 x 4.2 + 18 x 0
12 x 4.2
= 50.4
Sum of momentum after collision
(m1 + m2)v
= (12 + 18)v
= 30v
V = 50.4/30
= 1.7ms-1

1,782.

A car accelerate uniformly from rest at 4 ms-2. How far will it travel in the fifth complete second?

100 m

50 m

32 m

18 m

View answer & explanation

Correct answer is D

From S = ut + 1/2 at2
S = 1/2 ut2, since U = 0
Let S5 = distance in 5 seconds
= 1/2 x 4 x 52 = 2 x 25 = 50m
S4 = distance covered in 4 seconds
= 1/2 x 4 x 42 = 2 x 16 = 32m
Distance covered in the fifth second
S5 - S4 = 50m - 32m = 18m

1,783.

The value of x, y and z respectively in the expression M\(^x\) L\(^y\)T \(^z\) for the universal gravitational constant G are?

2, -3, -2

-1, 3, -2

-1, 2, -3

-2, -1, 3

View answer & explanation

Correct answer is B

F ∝	M1M2
	__r2

=> F =	GM1M2
	__r2

where F is the force of attraction
M1 and M2 are masses of the two bodies, r is the distance between them, and G is the universal gravitational constant

∴ G =	_Fr2
	M1M2

∴ dimension for force, F =	ML	= MLT-2
	T.T

The dimension for G =	MLT-2L2	=	L3T-2
	_MM		_M

= M-1L3T-2
In comparison with MxLyTz, the values for x, y and z = -1, 3, -2

1,784.

14	N +	4	He →	17	O + X
7		2		8

In the equation above, the particles X is?

a proton

a neutron

an α - particle

a β - particle

View answer & explanation

Correct answer is A

for the conversion of mass 14 + 4 = 17 + x
∴ X = 18 - 17 = 1
for conservation pf charge:7 + 2 = 8 + y
y = 9 - 8 = 1
Thus, the element is 11X which is equal to hydrogen nucleus ( proton)