In july 2005, nasa’s "deep impact" mission crashed a 372-kg probe directly onto the surface of the comet tempel 1, hitting the surface at 37,000 km/h. the original speed of the comet at that time was about 40,000 km/h, and its mass was estimated to be in the range (0.10−2.5)×1014kg. use the smallest value of the estimated mass.
(a) what change in the comet’s velocity did this collision produce? would this change be noticeable?
(b) suppose this comet were to hit the earth and fuse with it. by how much would it change our planet’s velocity? would this change be noticeable? (the mass of the earth is 5.97×1024kg.)

a)4.0*10^4 km/hr approx (the change will not be noticeable) b) 110050km/hr approx (this change will not be noticeable)

Explanation:

Using the law of conservation of momentum;

M1u1 + M2u2 = V(M1+M2)

Where M1 = mass of the probe = 372kg

M2 = mass of the comet = 0.1*10^14kg using the smallest value

U1 = velocity of the probe = 37000km/hr

U2 = velocity of the comet = 40000km/hr

V = final velocity of the comet with the probe

Make V subject of the formula = M1U1+M2U2/ (M1+M2) substitute the values gives

(372*37000)+(0.1*10^14*40000) /(372+0.1*10^14)

= 3.99999*10^4 approx 4.0*10^4 km/hr

The mass is too small compare to the mass of the comet to give a noticeable effect.

b) using the same law

M1 = mass of the comet plus probe aprox mass of comet = 0.1*10^14kg

M2 = mass of the earth = 5.97*10^24kg

U1 = initial velocity of the comet plus probe

U2 = initial velocity of the earth = 110,000km/hr

V = final velocity of the earth plus comet and probe)

V= M1U1+M2U2/(M1+M2)

V= (0.1*10^14*40000)+(5.97*10^24*110000)/(0.1*10^14+5.97*10^24) = 110050

It has an increment of 50km/hr

In fraction it = 50/110000 = 0.00045; it is too small to be noticed

the entire statement is true and correct.  

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