Question

A block of mass m = 4.4 kg, moving on frictionless surface with a speed vi = 9.2 m/s, makes a sudden perfectly elastic collision with a second block of mass M, as shown in the figure. The second block is originally at rest. Just after the collision, the 4.4-kg block recoils with a speed of vf = 2.5 m/s. What is the mass M of the second block? 7.7 kg6.2 kg10 kg2.2 kg14 kg

Answer 1

7.7kg

Step-by-step explanation:

Answer 2

Given data:

The mass of block 1 is m=4.4 kg.

The mass of block 2 is M.

The speed of block is vi=9.2 m/s.

The final speed of block 1 is vf=-2.5 m/s (negative because the block is moving in opposite direction after collision).

The expression for the final speed in elastic collision is given by,

`v_f=(m-M)/(m+M)v_i`

Substitute the given values in above equation,

`\begin{gathered} -2.5=(4.4-M)/(4.4+M)9.2 \\ -11-2.5M=40.48-9.2M \\ 6.7M=51.48 \\ M=7.7\text{ kg} \end{gathered}`

Thus, the value of mass M is 7.7 kg.

Final answer: 7.7 kg.