A 3 kg block, initially moving due east at 6 m/s, is acted upon by an impulse having magnitude 6 Ns and direction due west. The final velocity of the block is
A) 8 m/s east.
B) 4 m/s east.
C) 4 m/s west.
D) 8 m/s west.

Question

Quizplus · Accepted Answer

The impulse changes the momentum of the block. The impulse is given by $I = F\Delta t = \Delta p = m\Delta v$, where $I$ is the impulse, $F$ is the force, $\Delta t$ is the time interval, $\Delta p$ is the change in momentum, $m$ is the mass, and $\Delta v$ is the change in velocity. Given $I = 6\, \text{Ns}$ west, and the initial velocity is $6\, \text{m/s}$ east, the change in velocity can be calculated as $\Delta v = \frac{I}{m} = \frac{6}{3} = 2\, \text{m/s}$ to the west. Therefore, the final velocity is $6\, \text{m/s} - 2\, \text{m/s} = 4\, \text{m/s}$ east.

A 3 Kg Block, Initially Moving Due East at 6