1.

Showthatdl/dt=τ,whereListheangularmomentumandτisthetorque.​

Answer»

Answer:

GIVEN,

MASS=1.67×10

27

kg, Velocty of proton is 7.0×10

6

m/s Velocity of neutron is 4.0×10

6

m/s

So, BY the conservation of momentum, let m be the mass of the proton, so 2M is the mass of the deuteron.

Initially, the momentum of the proton is −m×7×10

6

kgm/s

Initial momentum of the neutron is +m×4×10

6

kgm/s

so, TOTAL momentum is −m×10

6

kgm/s

Now, after the ATTACHMENT of the mass 2m,

Let v be the velocity of the deuteron,

Then 2mv=−m×3×10

6

Canceling m then e get,

v=−1.5×10

6

m/s


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