Question

If the momentum p for a particle moving with velocity v=12i+yj+22k is p=xi+12j+33k. Find mass m,X,y for p=mv

Answer 1

The values of m, x and y are as following

m = 3/2 units

x = 18

y = 8

Explanation:

Given

The momentum $\vec p$ of the particle

$\vec p=x\hat i+12\hat j+33\hat k$

The velocity of the particle is

$\vec v=12\hat i+y\hat j+22\hat k$

If the mass of the particle is m then

We know that momentum is given by

$\vec p=m\vec v$

$\implies x\hat i+12\hat j+33\hat k= m(12\hat i+y\hat j+22\hat k)$

$\implies x\hat i+12\hat j+33\hat k = 12m\hat i+ym\hat j+22m\hat k$

On comparing the LHS and RHS

$x=12m$  ............. (1)

$12=ym$   ...............(2)

$33=22m$

$\implies m=\frac{33}{22}$

$\implies m=\frac{3}{2}$ unit

Therefore, from (1)

$x=12\times\frac{3}{2}$

$\implies x=18$

From eq (2)

$12=y\times\frac{3}{2}$

$\implies y=8$

Hope this answer is helpful.

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Answer 2

Answer:

•mass = (3/2) kg

• (x)= 18

• (y)= 8

Explanation:

(i) Given;

P = xi+12j+33k

v = 12i+yj+22k

(ii) We know; P=mv

(xi+12j+33k)=(m)(12i+yj+22k)

(xi+12j+33k)=(12mi+myj+22mk)

(iii)Comparing;

•33=22m (k)

(m)=(3/2) kg

•(x)=(12m)=(12)(3/2)=(18)

•(y)=(12/m)=(12)(2/3)=(8)

Hope It Helps!