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
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
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