Science, asked by anuj6185, 1 year ago

two hockey players player a of mass 50 gram is moving with a velocity of 4 metre per second and another one B belonging to opposite team with mass 60 kg is moving with velocity 3 metre per second and Tangled while chasing and fell down find the velocity with which they fall down​

Answers

Answered by Ug04
23

This can be solved using law of conservation of momentum

M¹=50kg.

( it should be in kg not in gm, But I think you entered wrong unit as a player can't have weight in grams **. )

M²=60kg

U¹=4m/sec

U²=3m/sec. V(1)+(2)=?

By law of conservation of momentum

(M¹×U¹) + (M²×U²)= (M¹×V¹) + (M²×V²)

50×4 + 60 × 3 = (M¹+M²) × V

{As they tangled there final velocity gets the same and mass adds up .*}

200 + 180 = 110 × V

380/110=V

V=3.454m/sec

Answered by Anonymous
1

Given,

mA = 50 g = 0.05 kg

uA = 4 m/s

mB = 60 kg

uB = 3 m/s

Initial momentumA = mAuA

= 0.05 × 4

= 0.2 kg m/s

Initial momentumB = mBuB

= 60 × 3

= 180 kg m/s

So,

Total initial momentum

= 0.2 + 180

= 180.2 kg m/s ...(i)

Final momentum

= (mA + mB)v

= (0.05 + 60)v

= 60.05v ...(ii)

According to the law of conservation of momentum,

180.2 = 60.05v

v = 180.2 /60.05

= 3.0 m/s

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