Question

A bullet with mass m = 5.21 g is moving horizontally with a speed v=367 m/s when it strikes a block of hardened steel with mass M = 14.8 kg(initially at rest). The bullet bounces o the block in a perfectly elastic collision.
(a) What is the speed (m/s) of the block immediately after the collision?
(b) What is the impulse (kg m/s) exerted on the block?
(c) What is the nal kinetic energy (J) of the block?
(d) How much work (J) did the bullet do on the block?
(e) What was the change in kinetic energy (J) of the bullet?
(f) How much work (J) was done on the bullet?

Answer 1

I will be happy to check your work.

Answer 2

m1 TIMES v1 = m2 TIMES v2
.0521 X 367= 14.8 X ?
.0521 X 367
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14.8
the division should be the answer of wat u seek

Answer 3

ytujytu