Question

A 50kg man is riding on a 40kg cart traveling at a speed of 4m/s. he jumps off with zero horizontal speed relative to the ground. What is the resulting speed of the cart?

Answer 1

Given:

A 50kg man is riding on a 40kg cart traveling at a speed of 4m/s. He jumps off with zero horizontal speed relative to the ground.

To find:

Resulting speed of cart.

Calculation:

Let resulting velocity of cart be v;

Let mass of man be m1 and that of cart be m2.

Applying Conservation of Momentum:

$\sf{\therefore \: (m1)(u1) + (m2)(u2) = (m1)(v1) + (m2)(v2)}$

$\sf{ = > \: (50)(4) + (40)(4) = (50)(0) + (40)(v)}$

$\sf{ = > \:200 + 160= 0 + 40v}$

$\sf{ = > \:360= 0 + 40v}$

$\sf{ = > \:360= 40v}$

$\sf{ = > \: v = \dfrac{360}{40} }$

$\sf{ = > \: v = 9 \: m {s}^{ - 1} }$

So, final answer is:

$\boxed{ \large{ \green{\rm{ \: v = 9 \: m {s}^{ - 1} }}}}$

Answer 2

Answer:

given =  

m1 = 50Kg

m2 = 40Kg

v1 = 4m/s

to find = resulting speed of the cart .i.e. v2 =?

solution=

Applying Conservation of Momentum:  

m1 u1 + m2 u2 = m1 v1 + m2 v2

50 ×4 + 40×4 = 50 ×0 + 40 × V2

200 + 160 =  0 + 40v2

360 = 40V2

V2 = 360 / 40

V2 =  9 ms⁻