if a body of mass 2kg travels with 2 m/s and other body of mass 2kg travels with 1m/s then find v1 and v2 in given statement​

Correct option is

A

both particles move together with velocity (2/3) m/s

B

the momentum of system is 2 kg m/s throughtout

D

the loss of KE of system is (49/3) J

Given : m

1

=2kg m

2

=1kg u

1

=3 m/s u

2

=4 m/s

Let the common velocity of the combined body be V

Mass of combined body M=2+1=3kg

Applying conservation of momentum :

2×3−1×4=3×V ⟹V=

3

2

m/s

Momentum of the system P=2×3−1×4=2 kg m/s

Loss in kinetic energy of the system due to collision ΔK.E=K.E

i

−K.E

f

ΔK.E=

2

1

(2)×3

2

+

2

1

(1)×4

2

−

2

1

(3)×(

3

2

)

2

⟹ΔK.E=

3

49

J

ill do it for 100, also this isnt 80 its 40

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases . this process does not change the mass number.

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as -decay. in this the mass number remains same.

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.