Question

d. Equal or less than that of the bullet 39. If the mass of the body is made three times and the velocity becomes double then the kinetic energy will increase? 6 times 12 In b. 12 times a. tor (B) (iz)? C. 24 times 2 (2011) d. 18 times RELATION BETEEN K.E & MOMENTUM 4 A particle of mass m has momentum P, its​

Answer 1

Answer:

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Explanation:

$so \: we \: know \: that \\ kinetic \: energy \: (K.E) = \frac{1}{2} mv {}^{2} \: \: \: \: \: \: \: \: \: (1) \\ \\ so \: but \: here \: it \: is \: given \: that \\ new \: mass \: of \: body \: (m1) = 3 .m \\ new \: velocity \: (v1) = 2.v \\ \\ so \: thus \: then \: \\ new \: kinetic \: energy \: is \: K.E \: ' = \frac{1}{2} m1.v1 {}^{2} \\ \\ = \frac{1}{2} \times 3m \times (2v) {}^{2} \\ \\ = \frac{1}{2} \times 3m \times 4v {}^{2} \\ = 3m \times 2v {}^{2} \\ = 6mv {}^{2} \\ \\ K.E \: '= 6mv {}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (2)$

$so \: now \: dividing \: (1) \: and \: (2) \\ we \: get \\ \frac{K.E}{K.E \: '} = \frac{ \frac{mv {}^{2} }{2} }{6mv {}^{2} } \\ \\ = \frac{mv {}^{2} }{2 \times 6mv {}^{2} } \\ \\ = \frac{1}{12} \\ \\ ie \: \: 12.K.E = K.E \: ' \\ \\ s o\: hence \: kinetic \: energy \: would \: get \: increased \: \: 12.times$