Question

A(n) 6 kg object moving with a speed of 8.6 m/s collides with a(n) 17 kg object moving with a velocity of 11 m/s in a direction 20◦ from the initial direction of motion of the 6 kg object. 8.6 m/s 11 m/s 17 kg 6 kg 20◦ What is the speed of the two objects after the collision if they remain stuck together? Answer in units of m/s.

Answer 1

Answer:

The speed of the two objects after the collision if they remain stuck together is 10.26 m/s.

Explanation:

Given that,

Mass of object = 6 kg

Speed of object = 8.6 m/s

Other object = 17 kg

Velocity = 11 m/s

Angle = 20°

We need to calculate the speed of the two objects after the collision if they remain stuck together

Using conservation of momentum

$m_{1}v_{1}+m_{2}v_{2}=Mv$

Put the value into the formula

On x axis,

$6\times8.6+17\times11\cos20=23v\cos\theta$...(I)

On y axis,

$0+17\times11\sin20=23v\sin\theta$....(II)

Squaring and addin equation (I) and (II)

$v=\dfrac{(6\times8.6+17\times11\cos20)^2+(17\times11\sin20)^2}{23^2}$

$v=\sqrt{105.41}\ m/s$

$v=10.26\ m/s$

Hence, The speed of the two objects after the collision if they remain stuck together is 10.26 m/s.

Answer 2

The speed of the two objects after collision is 9.88 m/s.

The given parameters;

• mass of the first object, m₁ = 6 kg
• initial speed of the first object, u₁ = 8.6 m/s
• mass of the second object, m₂ = 17 kg
• initial velocity of the second object, u₂ = 11 m/s at 20⁰

The speed of the two objects after collision is calculated by applying the principle of conservation of linear momentum for inelastic collision as follows;

$m_1u_1 + m_2 u_2 = v(m_1 + m_2)\\\\6(8.6) \ + \ 17(11 \times cos20)= v(6 + 17)\\\\227.32 = 23v\\\\v = \frac{227.32}{23} \\\\v = 9.88 \ m/s$

Thus, the speed of the two objects after collision is 9.88 m/s.

Learn more about inelastic collision here: https://brainly.com/question/7694106