Question

A 42-g firecracker is at rest at the origin when it explodes into three pieces. The first, with mass 12 g, moves along the x-axis at 35 m/s. The second, with mass 21 g, moves along the y-axis at 29 m/s. Find the velocity of the third piece.

Answer 1

`overrightarrow{v_(3)}=-46.67\widehat{i}-67.67\widehat{j}`

Step-by-step explanation:

M = 42 g

U = 0

m1 = 12 g

vi = 35 m/s along X axis = 35 i

m2 = 21 g

v2 = 29 m/s along Y axis = 29 j

Let v3 be the velocity of third part. m3 = M - m1 - m2 = 42 g - 12 g - 21 g = 9 g

Use conservation of momentum

`M * \overrightarrow{U}= m_(1)\overrightarrow{v_(1)}+m_(2)\overrightarrow{v_(2)}+m_(3)\overrightarrow{v_(3)}`

`0=12* 35\widehat{i}+21* 29\widehat{j}+9*\overrightarrow{v_(3)}`

`overrightarrow{v_(3)}=-46.67\widehat{i}-67.67\widehat{j}`

Answer 2

u₃ = - 46.66 i -67.66 j

Step-by-step explanation:

Given that

M= 42 g

Initial speed ,u= 0 m/s

m₁ = 12 g ,u₁=35 i m/s

m₂=21 g,u₂=29 j m/s

Lets take mass of the third mass = m₃

The speed of the third mass = u₃ m/s

From mass conservation

M= m₁+ m₂ + m₃

42 = 12 + 21 + m₃

m₃= 9 g

There is no any external force that is why linear momentum will be conserve

M u = m₁ u₁+m₂u₂+m₃u₃

42 x 0 = 35 i x 12 + 21 j x 29 + 9 u₃

9 u₃ = -420 i - 609 j

u₃ = -46.66 i -67.66 j

Therefore the speed of the third block will be

u₃ = - 46.66 i -67.66 j