Answer:(d) 3750J
From the given data
The initial momentum is
2*50-4*25 =0
Then after collision they stick together then the final velocity is zero
Now the loss in kinetic energy is = (1/2)(2)(50)2+(1/2)(4)(25)2 =3750J
13)
Answer is (b)
First velocity of the ball just before impact is (v1) =10m/s
The velocity just after touch the ground is v2 = √2*g*h2 =√2*9.81m/s2*2m=6.264m/s
Now the impulse is nothing but the change in momentum =mv1+mv2 =4(10)+4(6.264) =40+25.056 =65.056N.s
Need help with problem asap U. the greater c. the smaller momentum d. the greater speed...
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