A 1.6-kg ball is attached to the end of a 0.40-m string to form a pendulum. This pendulum is released from rest with the string horizontal. At the lowest point of its swing, when it is moving horizontally, the ball collides with a 0.80-kg block initially at rest on a horizontal frictionless surface. The speed of the block just after the collision is 3.0 m/s. What is the speed of the ball just after the collision?a. 1.5 m/sb. 1.3 m/sc. 2.1 m/sd. 1.1 m/se. 1.7 m/s

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hamzaahmeds hamzaahmeds · Answer 1 · 2021-03-28T03:10:15+02:00

Answer:

a. 1.5 m/s

Explanation:

We will apply the law of conservation of energy in this situation between the initial position and the lowest point of the swing:

[tex]m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}\\[/tex]

where,

m₁ = mass of ball = 1.6 kg

m₂= mass of block = 0.8 kg

u₁ = initial speed of ball = 0 m/s

u₂ = initial speed of block = 0 m/s

v₁ = final speed of ball = ?

v₂ = final speed of block = 3 m/s

Therefore,

[tex](1.6\ kg)(0\ m/s)+(0.8\ kg)(0\ m/s)=(1.6\ kg)(v_{1})+(0.8\ kg)(3\ m/s)\\\\v_{1} = \frac{2.4\ N.s}{1.6\ kg}\\[/tex]

v₁ = 1.5 m/s

Therefore, the correct option is:

a. 1.5 m/s