Question

A 3.3 kg mass with an initial velocity of
v1 = aˆi, where a = 3.47 m/s collides with
and sticks to a 0.989 kg mass with an initial
velocity of v1 = bˆj, where b = −4.8 m/s.
Find the final speed of the composite mass.
Answer in units of m/s

Answer 1

I got several answers and they have all been wrong so far :(

Answer 2

conservation of momentum:

\[m_1 v_1 + m_2 v_2 = m_{total} v_{total} = (m_1 + m_2)v_{total}\]

\[3.3*(3.47 i) + 0.989*(-4.8 j) = (3.3 + 0.989)*v_{total}\]\[v_{total} = \frac{ 3.3*3.47 }{ 4.289 } i - \frac{ 0.989*4.8 }{ 4.289 }j \]\[\left| v_{total} \right| = \left| \sqrt{v_i^2 + v_j^2} \right|\]\[\theta = \tan^{-1}\left( \frac{ v_j }{ v_i } \right)\]

| this is absolute value |

lmk if you have questions ^_^

Answer 3

I got 3.09372 as the v total

Answer 4

But I inputted it and I got it wrong?

Answer 5

\[v_i \approx 2.6698\]\[v_j \approx 1.1068\]
\[v_tot \approx 2.89\]

Answer 6

wow thank you :)

Answer 7

glad i could help ^_^

Answer 8

could you try another one?

Answer 9

sure

Answer 10

A 7 kg steel ball strikes a wall with a speed
of 8.77 m/s at an angle of 59.4 with the
normal to the wall. It bounces off with the
same speed and angle

If the ball is in contact with the wall for
0.237 s, what is the magnitude of the average
force exerted on the ball by the wall?
Answer in units of N