The collision between a hammer and a nail can be considered to be approximately elastic.?

Question

The collision between a hammer and a nail can be considered to be approximately elastic. Estimate the kinetic energy acquired by a 15 g nail when it is struck by a 550 g hammer moving with an initial speed of 3.5 m/s.

This is a practice problem from my physics book. So I didn't copy the question incorrectly. It indeed says elastic...

Thanks. Those are the velocities I got but I didn't understand what they were asking for as far as KE.

Answer 1

Obviously the writer of the problem never hammered a nail. See the ref. Assuming no friction, the following are the results.
Notation is: 1, 2, CM, Rel = hammer, nail, system, relative (body 1 relative to body 2); (i), (f) = initial, final; V = velocity, M = mass, CR = coefficient of restitution (=1 when elastic), KE = kinetic energy.
VCM = (M1V1+M2V2)/(M1+M2) = 3.4070796460177
VRel(i) = V1(i)-V2(i) = 3.5
VRel(f) = -CR*VRel(i) = -3.5
V1(f) = VCM+VRel(f)*M2/(M1+M2) = 3.3141592920354
V2(f) = VCM-VRel(f)*M1/(M1+M2) = 6.8141592920354
KE1(f) = M1*V1(f)^2/2 = 3.02050424857076
KE2(f) = M2*V2(f)^2/2 = 0.348245751429243
KE2(f) is the answer to the question.
EDIT: Pearl used 0.15 kg as the nail mass, not the correct value of 0.015 kg. Also, my solution is fully elastic (and as mentioned above, fully unrealistic).

Answer 2

Vn = 2e6e251890bde37c6fa9136d41d04a8[e6e251890bde37c6fa9136d41d04a8e6e251890bde37c6fa9136d41d04a8e6e251890bde37c6fa9136d41d04a8e6e251890bde37c6fa9136d41d04a8e6e251890bde37c6fa9136d41d04a8e6e251890bde37c6fa9136d41d04a8e6e251890bde37c6fa9136d41d04a8e6e251890bde37c6fa9136d41d04a8e6e251890bde37c6fa9136d41d04a8e6e251890bde37c6fa9136d41d04a8e6e251890bde37c6fa9136d41d04a8e6e251890bde37c6fa9136d41d04a8e6e251890bde37c6fa9136d41d04a8e6e251890bde37c6fa9136d41d04a8] = 7.8 me6e251890bde37c6fa9136d41d04a8s KE = ½mVn² = .e6e251890bde37c6fa9136d41d04a8e6e251890bde37c6fa9136d41d04a8.e6e251890bde37c6fa9136d41d04a8e6e251890bde37c6fa9136d41d04a8e6e251890bde37c6fa9136d41d04a8e6e251890bde37c6fa9136d41d04a87.8² = .e6e251890bde37c6fa9136d41d04a82e6e251890bde37c6fa9136d41d04a89 J a pair of the solutions above assume each and all the momentum of the hammer is transferred to the nail. rather, the hammer nevertheless has momentum after the impression. In an elastic collision, the middle of mass of the gadget keeps a persevering with velocity, and the velocities of the two bodies relative to the CM are the comparable with the signs and indications reversed till now and after. interior the case of one physique being initially at relax, its Vr = Vc = Pie6e251890bde37c6fa9136d41d04a8?M, so its very final velocity would be Vc 550*4/(550+14) Vr = 2e6e251890bde37c6fa9136d41d04a8Vc.........

Answer 3

http://hyperphysics.phy-astr.gsu.edu/hbase/colsta.html#c5
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The head on elastic collision relation ships:
V1' = {[m1 -m2 ] / [m1 +m2]} V1 :

V2" = {2m1 / [m1 +m2]} V1.
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Kirchwey had used the formula for in elastic collision

Kirchwey had used the formula for inelastic collision.

m1 = 0.15 kg,
m2 = 0.55 kg,
v1= 0,
V2 = 3.5 m/s
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v1' =? v2" = ?
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Substituting this in the formula or writing the values in the web site we get

v1' = 5.5 m/s and
v2' = 2m/s.

Both move in the original direction of the hammer.

Kinetic energy of the nail = 0.5 m V2' ^2 = 0.226875 J.
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