Question

15. An object of mass 1 kg travelling in a straight line with a velocity
of 10 msl collides with, and sticks to, a stationary wooden
block of mass 5 kg. Then they both move off together in the
same straight line. Calculate the total momentum just before
the impact and just after the impact. Also, calculate the velocity
of the combined object.​

Answer 1

ANSWER

Given,

m=1kg,u1=10ms−1,M=5kgandu2=0ms−1

From conservation of momentum

Initial momentum = final momentum

mu1+Mu2=(m+M)v

1×10+5×0=(1+5)v

10=6v

Hence, velocity of both isv=35ms−1

Total momentum before is 10kgms−1

From momentum conservation, total momentum after is 10kgms−1

Answer 2

Given :-

$\textsf{• Mass of object A = 1 kg}$

$\textsf{• Velocity of object A = 10 m/s }$

$\textsf{• Mass of object B = 5 kg}$

$\textsf{• Velocity of object B = 0 m/s (It is stationary)}$

To find :-

$\textsf{• Total momentum before impact }$

$\textsf{• Total momentum after impact }$

$\textsf{• Velocity of combined object }$

According to the question,

$\textsf{•➝ Momentum of object A = m × v}$

$\textsf{•➝ Momentum of object A = 1 × 10 }$

$\textsf{•➝ Momentum of object A = 10 kg m/s }$

Solution :-

$\textsf{•➝ Momentum of object B = m × v}$

$\textsf{•➝ Momentum of object B = 5 × 0}$

$\textsf{•➝ Momentum of object B = 0 kg m/s }$

$\textsf{.°. Total momentum before collision = 10 kg m/s + 0 kg m/s }$

$\textsf{•➝ 10 kg m/s }$

Now,

$\textsf{•➝ Total mass = 1 kg + 5 kg }$

$\textsf{•➝ Total mass = 6 kg}$

$\textsf{•➝ Momentum after collision = m × v}$

$\textsf{•➝ Momentum after collision = 6 × v}$

$\textsf{•➝ Momentum after collision = 6v }$

So,

$\textsf{•➝ Momentum before collision = Momentum after collision }$

$\textsf{•➝ 10 = 6 × v}$

$\textsf{•➝ 10 ÷ 6 = v}$

$\textsf{•➝ 1.67 = v}$

$\textsf{So, velocity of combined object is 1.67 m/s.}$