A block of mass m = 4.4 kg,moving on frictionless surface with a speed $v _ { \mathrm { i } } = 9.2 \mathrm {~m} / \mathrm { s }$ 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 $v _ { \mathrm { f } } = 2.5 \mathrm {~m} / \mathrm { s }$ What is the mass M of the second block?
A)7.7 kg
B)0 kg
C)0.0 kg
D)0.7 kg

Question

A block of mass m = 4.4 kg,moving on frictionless surface with a speed $v _ { \mathrm { i } } = 9.2 \mathrm {~m} / \mathrm { s }$ 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 $v _ { \mathrm { f } } = 2.5 \mathrm {~m} / \mathrm { s }$ What is the mass M of the second block? 
A)7.7 kg 
B)0 kg 
C)0.0 kg 
D)0.7 kg

Quizplus · Accepted Answer

Let the mass of the second block be M.

Conservation of momentum:
Before the collision: m×vi = M×0 (second block is at rest)
After the collision: m×vf = M×v (where v is the velocity of the second block after the collision)

Conservation of kinetic energy:
Before the collision: ½m×vi² + 0 = ½M×0² + 0 (second block is at rest)
After the collision: ½m×vf² + ½M×v² = ½m×vi²

Solving both equations for M, we get:

M = (m×vi²)/(2×v² - vi²) = (4.4×9.2²)/(2×2.5² - 9.2²) ≈ 7.7 kg

Therefore, the answer is A) 7.7 kg.

A Block of Mass M = 4 vi=9.2 m/sv _ { \mathrm { i } } = 9.2 \mathrm {~m} / \mathrm { s }vi​=9.2 m/s

A Block of Mass M = 4 $v _ { \mathrm { i } } = 9.2 \mathrm {~m} / \mathrm { s }$