Asked by Anonymous
A 2,000-kg car moving east at 10.0 m/s collides with a 3,000-kg car moving north. The cars stick together and move as a unit after the collision, at an angle of 39.0° north of east and at a speed of 5.15 m/s. Find the speed of the 3,000-kg car before the collision.
Answers
Answered by
Steve
keep the momentum the same
2000*10<b>i</b> + 3000v<b>j</b> = 5000*5.15(cos39°<b>i</b> + sin39°<b>j</b>)
Now, 5000*5.15*cos39° = 20000
so, 3000v = 5000*5.15*sin39° = 16196
v = 5.40
2000*10<b>i</b> + 3000v<b>j</b> = 5000*5.15(cos39°<b>i</b> + sin39°<b>j</b>)
Now, 5000*5.15*cos39° = 20000
so, 3000v = 5000*5.15*sin39° = 16196
v = 5.40
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