Respuesta :
Answer:
u₃ = - 46.66 i -67.66 j
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
Answer:
[tex]overrightarrow{v_{3}}=-46.67\widehat{i}-67.67\widehat{j}[/tex]
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
[tex]M \times \overrightarrow{U}= m_{1}\overrightarrow{v_{1}}+m_{2}\overrightarrow{v_{2}}+m_{3}\overrightarrow{v_{3}}[/tex]
[tex]0=12\times 35\widehat{i}+21\times 29\widehat{j}+9\times\overrightarrow{v_{3}}[/tex]
[tex]overrightarrow{v_{3}}=-46.67\widehat{i}-67.67\widehat{j}[/tex]
