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Physics

Grade-8

Easy

Question

A 3,000 kg truck moving at +10 $fraction numerator m over denominator s end fraction$ hits a 1,000 kg parked car which moves off at +15 $fraction numerator m over denominator s end fraction$ . What is the velocity of the truck?

5 $\frac{m}{s}$
50 $\frac{m}{s}$
2 $\frac{m}{s}$
0.2 $\frac{m}{s}$

hint Hint:

The formula for the law of conservation of momentum is: m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2.

The correct answer is: 5 $fraction numerator m over denominator s end fraction$

The formula for the law of conservation of momentum is: m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2.

parked car means u2=0

let velocity of truck is v1

$3000 cross times 10 plus 1000 cross times 0 equals 1000 cross times 15 plus 3000 v 1 30000 minus 15000 equals 3000 v 1 15000 equals 3000 v 1 v 1 equals 5 space m divided by s$

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