Question

Two blocks of mass 0.1 kg and 0.2 kg approach each other on a horizontal plane at velocities of 0.4 and 1 m/s respectively. If the blocks collide and remain together , calculate the joint velocity after collision.

Answer 1

Given,
M1=0.1
M2=0.2
V1=0.4
V2=1
M(combined mass)=0.3
V(final velocity)=?

Apply law of conservation of linear momentum
Initial linear momentum=Final linear momentum
M1V1-M2V2=MV
0.1*0.4-0.2*1=0.3V
0.04-0.2=0.3V
-0.16/0.3=V
-5.33=V
Velocity of resultant mass is in the direction of M2

Answer 2

Answer:

0.533m/s

Explanation:

Simce the bodies remain in contact after the collision, the collision in an inelastic collision.

In case of inelasetic collision, if a body with mass m1 initially moving with velocity v1 and undergoes collision with a body of mass m2 and velocity v2, then the resultant velocity (say v) is given by the expression below:

$v = \frac{m1 \times v1 + m2 \times v2}{m1 + m2}$

In this case m1=0.1kg, m2=0.2kg, v1=0.4m/s, v2=-1m/s, ( we have to consider the direction of velocities, so we are taking one to be positive and other to be negative.)then resultant velocity v is

$v = \frac{0.1kg \times 0.4m{ s}^{ - 1} - 0.2kg \times 1m {s}^{ - 1} }{0.1kg + 0.2kg} \\ = \frac{ 0.04kgm {s}^{ - 1 } - 0.2kgm {s}^{ - 1} }{0.3kg} \\ = \frac{-0.16kgm {s}^{ - 1} }{0.3kg} = -0.533m {s}^{ - 1}$

So, the final velocity of both the blocks are 0.533m/s (in the direction of initial motion of the 0.2kg block)

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