Two particles of masses 2kg and 3kg are projected horizontally in opposite directions from the top of a tower of height 39.2m with velocities 5ms- 1 and 10ms- 1 respectively. The horizontal range of the centre of mass of two particles is [ g=9.8ms- 2 ]

Question

Two particles of masses 2kg and 3kg are projected horizontally in opposite directions from the top of a tower of height 39.2m with velocities 5ms- 1 and 10ms- 1 respectively. The horizontal range of the centre of mass of two particles is [ g=9.8ms- 2 ] (A) 8√2 m in the direction of 2 kg (B) 8√2 m in the direction of 3 kg (C) √8 m in the direction of 2 kg (D) √8 m in the direction of 3 kg

Tardigrade Admin · Accepted Answer

Range of C.M =Vc m√(2 h/g) Vc m=(m1 v1 + m2 v2/m1 + m2) Vc o m = (- 2 × 5 + 3 × 10/2 + 3) = 4 m / s Range = 4 √(2 × 39.2/9.8) = 8 √2

Q. Two particles of masses $2 k g$ and $3 k g$ are projected horizontally in opposite directions from the top of a tower of height $39.2 m$ with velocities $5 m s^{- 1}$ and $10 m s^{- 1}$ respectively. The horizontal range of the centre of mass of two particles is [ $g = 9.8 m s^{- 2}$ ]

$82$ m in the direction of 2 kg

$82$ m in the direction of 3 kg

$8$ m in the direction of 2 kg

$8$ m in the direction of 3 kg

Range of C.M $= V_{c m} \frac{2 h}{g}$
$V_{c m} = \frac{m _{1} v _{1} + m _{2} v _{2}}{m _{1} + m _{2}}$
$V_{co m} = \frac{- 2 \times 5 + 3 \times 10}{2 + 3} = 4 m / s$
Range $= 4 \frac{2 \times 39.2}{9.8} = 82$

Q. Two particles of masses 2kg and 3kg are projected horizontally in opposite directions from the top of a tower of height 39.2m with velocities 5ms−1 and 10ms−1 respectively. The horizontal range of the centre of mass of two particles is [ g=9.8ms−2 ]

82​ m in the direction of 2 kg

82​ m in the direction of 3 kg

8​ m in the direction of 2 kg

8​ m in the direction of 3 kg