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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

A Block Of Mass M 44 Kg Moving On Frictionless Surface With A Speed Vi 92 Ms Makes A Sudden Perfectly Elastic Collision With A Second Block Of Mass M As Shown I class=

Sagot :

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,

[tex]v_f=\frac{m-M}{m+M}v_i[/tex]

Substitute the given values in above equation,

[tex]\begin{gathered} -2.5=\frac{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}[/tex]

Thus, the value of mass M is 7.7 kg.

Final answer: 7.7 kg.

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