
A mass of 2.9 kg is suspended from a string of length 50cm and is at rest. Another body of mass 100g, which is horizontally with a velocity of 150m/sec, sticks to it. What is the tension in the string when it makes an angle of with the vertical, the tension in the string is:
$(g = 10m/{s^2})$
A.)140 N
B.)135 N
C.)125 N
D.)90 N
Answer
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Hint: Momentum is conserved in inelastic collision, but one cannot track the kinetic energy through collision since some of it is converted to other forms of energy. Moreover an inelastic collision occurs when two objects collide and do not bounce away from each other.
Step By Step Answer:
Imagine a string having 2.9kg (given) and a mass of 0.1 strikes the string which is 100g (given) and the velocity of mass is given 150m/sec. and when the string is able to make an angle of ${60^\circ } $ with vertical according to the given question.
As the collision is perfectly inelastic, conserving momentum, we get,
$
0.1 \times 150 = (2.9 + 0.1) \times u \\
\Rightarrow u = 5m/s \\
\\
$
By applying conservation of energy,
$
(m + M)g(l(1 - \cos (60)) + \dfrac{1}{2}(m + M){v^2} = \dfrac{1}{2}(m + M){u^2} \\
\\
\\
$
By taking$\dfrac{1}{2}(m + M){v^2} $. out we get,
$
\dfrac{1}{2}(m + M){v^2} = \dfrac{1}{2}(m + M){u^2} - (m + M)gl(1 - \cos (60)) \\
(m + M){v^2} = 3 \times {5^2} - 2 \times 3 \times 10 \times 10.5 \times (1 - 0.5) \\
\Rightarrow (m + M){v^2} = 60 \\
T = (m + M)g\cos (\theta ) + (m + M){v^2}/l \\
{\text{ }} = 15 + 120 \\
T = 135N \\
$
Therefore, option (B) is the right answer.
Additional Information: TENSION: when two pulling forces are directed opposite to each other and stretch an object by trying to pull it apart that force is said to be tension.
Note: There is a difference between an elastic collision and inelastic collision; if the total momentum and total kinetic energy before and after the collision is same then it is said to be ecstatic collision whereas if some of energy is converted into any other form of energy let us say sound energy or thermal energy then it is said to be an inelastic collision.
Step By Step Answer:
Imagine a string having 2.9kg (given) and a mass of 0.1 strikes the string which is 100g (given) and the velocity of mass is given 150m/sec. and when the string is able to make an angle of ${60^\circ } $ with vertical according to the given question.
As the collision is perfectly inelastic, conserving momentum, we get,
$
0.1 \times 150 = (2.9 + 0.1) \times u \\
\Rightarrow u = 5m/s \\
\\
$
By applying conservation of energy,
$
(m + M)g(l(1 - \cos (60)) + \dfrac{1}{2}(m + M){v^2} = \dfrac{1}{2}(m + M){u^2} \\
\\
\\
$
By taking$\dfrac{1}{2}(m + M){v^2} $. out we get,
$
\dfrac{1}{2}(m + M){v^2} = \dfrac{1}{2}(m + M){u^2} - (m + M)gl(1 - \cos (60)) \\
(m + M){v^2} = 3 \times {5^2} - 2 \times 3 \times 10 \times 10.5 \times (1 - 0.5) \\
\Rightarrow (m + M){v^2} = 60 \\
T = (m + M)g\cos (\theta ) + (m + M){v^2}/l \\
{\text{ }} = 15 + 120 \\
T = 135N \\
$
Therefore, option (B) is the right answer.
Additional Information: TENSION: when two pulling forces are directed opposite to each other and stretch an object by trying to pull it apart that force is said to be tension.
Note: There is a difference between an elastic collision and inelastic collision; if the total momentum and total kinetic energy before and after the collision is same then it is said to be ecstatic collision whereas if some of energy is converted into any other form of energy let us say sound energy or thermal energy then it is said to be an inelastic collision.
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