A stationary object of mass 5.0\text{ kg} explodes into two fragments of masses 2.0\text{ … — JAMB UTME

A stationary object of mass $5.0\text{ kg}$ explodes into two fragments of masses $2.0\text{ kg}$ and $3.0\text{ kg}$ . If the $2.0\text{ kg}$ fragment moves due east at a velocity of $15\text{ m s}^{-1}$ , what is the velocity of the $3.0\text{ kg}$ fragment?

$10\text{ m s}^{-1}$ due westAnswer
B
$10\text{ m s}^{-1}$ due east
C
$15\text{ m s}^{-1}$ due west
D
$22.5\text{ m s}^{-1}$ due west

Answer

10\text{ m s}^{-1}

due west

Before the explosion, the system is stationary, giving an initial momentum of

0\text{ kg m s}^{-1}

. By the law of conservation of linear momentum, the total momentum after the explosion must also equal zero. Taking the eastward direction as positive, the

2.0\text{ kg}

fragment has a momentum of

+30\text{ kg m s}^{-1}

. To balance this, the

3.0\text{ kg}

fragment must possess a momentum of

-30\text{ kg m s}^{-1}

, yielding

v = -10\text{ m s}^{-1}

, which means a speed of

10\text{ m s}^{-1}

directed due west.

Step-by-Step Solution

Determine the initial momentum of the system

p_i = 0\text{ kg m s}^{-1}

The object is initially at rest.

Express final total linear momentum using vector direction (let east be positive)

p_f = m_1 v_1 + m_2 v_2 = (2.0 \times 15) + (3.0 \times v_2) = 30 + 3 v_2

Linear momentum is conserved in an isolated system.

Equate initial momentum to final momentum and solve for

v_2

0 = 30 + 3 v_2 \implies v_2 = -10\text{ m s}^{-1}

The negative sign indicates the direction is opposite to east, which is due west.

Key Concept

Conservation of Linear Momentum in Explosions

Estimated Time:1m 30s

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