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

ID: 1357872 • Letter: T

Question

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

Part A

Calculate the kinetic energy acquired by a 14-g nail when it is struck by a 530-g hammer moving with an initial speed of 4.8 m/s .

Express your answer using two significant figures.

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

Part A

Calculate the kinetic energy acquired by a 14-g nail when it is struck by a 530-g hammer moving with an initial speed of 4.8 m/s .

Express your answer using two significant figures.

Explanation / Answer

Solution: Let the speed of the hammer before collision be vhi = 4.8m/s

speed of the nail before collision be vni = 0 m/s (stationary)

speed of the hammer after collision be vhf

speed of the nail before collision be nf

mass of the hammer mh = 530 g = 0.53 kg

mass of the nail mn = 14 g = 0.014 kg

During the elastic collision, the total linear momentum is conserved, thus

Total momentum before collision = total momentum after collision

mh*vhi + mn*vni = mh*vhf + mn*vnf

mh*vhi = mh*vhf + mn*vnf                                                                                                   since vni = 0

mh*(vhi - vhf ) = mn*vnf                             ------------------------------------(1)

During the elastic collision the total kinetic energy is also conserved ,

Total kinetic energy before collision = total kinetic energy after collision

(1/2)mh*vhi2 + (1/2)mn*vni2 = (1/2)mh*vhf2 +(1/2) mn*vnf2

mh*vhi2 + 0 = mh*vhf2 + mn*vnf2

mh*(vhi2 - vhf2 )= mn*vnf2

mh*(vhi + vhf ) *(vhi - vhf )= mn*vnf2 ------------------------------------------------------(2)

divide equation(2) by (1) we get

(vhi + vhf ) = vnf

vhf = vnf - vhi

Putting this vhf in equation (1) we get

mh*(vhi - vnf + vhi ) = mn*vnf

mh*2vhi – mh*vnf   = mn*vnf

mh*2vhi   = mn*vnf + mh*vnf

mh*2vhi = vnf*(mn +mh)

vnf = 2mhvhi­/(mn +mh)

Thus

vnf = (2*0.53kg*4.8m/s) / (0.014+0.53)kg

vnf = 9.3529 m/s

Thus the kinetic energy of the nail is

KEnf = (1/2) mn*vnf2

KEnf = (1/2)*0.014kg*(9.3529m/s)2

KEnf = 0.6123 J

Thus the kinetic energy acquired by the nail is 0.61 J