Question

5.
A mass of 3.9 kg suspended from a string of length 0.5m is at rest. Another body of mass 0.1kg moving
horizontally with a velocity 200 ms' strikes and sticks to it. The tension in the string when it makes 60%
with the vertical is (g = 10 ms)​

Answer 1

Answer:

Tension in the string is 180 N at given position

Explanation:

As we know that the mass of 3.9 kg is suspended from the string and another mass makes perfect inelastic collision with it

So we will have

$mv = (m + M) v_f$

so we have

$0.1 (200) = (0.1 + 3.9) v_f$

$v_f = 5 m/s$

now when string makes 60 degree with the vertical the speed is given as

$v_f^2 = v_i^2 - 2gL(1 - cos60)$

$v_f^2 = 5^2 - 2(10)(0.5)(1 - 0.5)$

$v_f = 4.47 m/s$

Now at that position we can write the force equation as

$T - mg cos60 = \frac{mv^2}{L}$

$T = 4(10)cos60 + \frac{4(20)}{0.5}$

$T = 180 N$

#Learn

Topic : vertical circular motion

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