Question

Car A is 3000N moves at velocity 3 m/s east. A car B 2500 N approaches car A at velocity 4m/s west. Determine the velocity of car A if (a) the two cars get stuck together after collision; (b) car B moves 5m/s west; (c) car B moves 3m/s east

Answer 1

Given:

Car A (of weight 3000N) moves at velocity 3 m/s east. A car B (of weight 2500 N) approaches car A at velocity 4m/s west.

To find:

velocity of car A if

• the two cars get stuck together after collision
• car B moves 5m/s west
• car B moves 3m/s east

Calculation:

In all the cases , since no external force has been applied we can easily use the principle of Conservation of Momentum to solve this problem.

Mass of car A = 3000/10 = 300 kg

Mass of car B = 2500/10 = 250 kg

[ Let due east be positive ]

1st part :

$(m1)(v1) + (m2)(v2) = (m1 + m2)(v)$

$= > (300 \times 3) - (250 \times 4) = (300 + 250)v$

$= > - 100 = 550v$

$= > v = - 0.18 \: m {s}^{ - 1}$

So the combination velocity of both cars will be 0.18 m/s towards west.

Part 2:

$(m1)(u1) + (m2)(u2) = (m1)(v1) + (m2)(v2)$

$= >( 300 \times 3) - (250 \times 4) = (300 \times v1) - (250 \times 5)$

$= > - 100 = (300 \times v1) - 1250$

$= > 1150 = (300 \times v1)$

$= > v1 = 3.83 \: m {s}^{ - 1}$

So car A will have velocity of 3.83 m/s due east.

Part 3:

$(m1)(u1) + (m2)(u2) = (m1)(v1) + (m2)(v2)$

$= >( 300 \times 3) - (250 \times 4) = (300 \times v1) + (250 \times 3)$

$= > - 100 = (300 \times v1) + 750$

$= > 300(v1) = - 850$

$= > v1 = - 2.83 \: m {s}^{ - 1}$

So car A will move due west with velocity of 2.83 m/s.